Course 421: Michaelmas Term 2002 Topological Spaces and the Fundamental

Group

David R. Wilkins

Copyright \begin{matrix}c \\ ◯ \end{matrix} David R. Wilkins 1988–2002

Contents

1 Topological Spaces 3 1.1 The Concept of Continuity . . . . . . . . . . . . . . . . . . . . 3 1.2 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . 3 1.3 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . 4 1.4 Open Sets in Metric Spaces . . . . . . . . . . . . . . . . . . . 5 1.5 Further Examples of Topological Spaces . . . . . . . . . . . . 7 1.6 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.7 Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.8 Subspace Topologies . . . . . . . . . . . . . . . . . . . . . . . 8 1.9 Continuous Functions between Topological Spaces . . . . . . . 10 1.10 Continuous Functions between Metric Spaces . . . . . . . . . . 10 1.11 A Criterion for Continuity . . . . . . . . . . . . . . . . . . . . 11 1.12 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.13 Neighbourhoods, Closures and Interiors . . . . . . . . . . . . . 13 1.14 Product Topologies . . . . . . . . . . . . . . . . . . . . . . . . 14 1.15 Identification Maps and Quotient Topologies . . . . . . . . . . 18 1.16 Compact Topological Spaces . . . . . . . . . . . . . . . . . . . 19 1.17 The Lebesgue Lemma and Uniform Continuity . . . . . . . . . 25 1.18 Connected Topological Spaces . . . . . . . . . . . . . . . . . . 27

2 Homotopies and Covering Maps 32 2.1 Homotopies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.2 Covering Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3 Path Lifting and the Monodromy Theorem . . . . . . . . . . . 34 3 The Fundamental Group 36

3.1 The Fundamental Group of a Topological Space . . . . . . . . 36 3.2 Simply-Connected Topological Spaces . . . . . . . . . . . . . . 39 3.3 The Fundamental Group of the Circle . . . . . . . . . . . . . . 41

4 Simplicial Complexes 43 4.1 Geometrical Independence . . . . . . . . . . . . . . . . . . . . 43 4.2 Simplicial Complexes in Euclidean Spaces . . . . . . . . . . . 45 4.3 Simplicial Maps . . . . . . . . . . . . . . . . . . . . . . . . . . 47 4.4 Barycentric Subdivision of a Simplicial Complex . . . . . . . . 48 4.5 The Simplicial Approximation Theorem . . . . . . . . . . . . . 51 4.6 The Brouwer Fixed Point Theorem . . . . . . . . . . . . . . . 53

5 An Application of Topology to Mathematical Economics 55 5.1 The Existence of Equilibria in an Exchange Economy . . . . . 55

1 Topological Spaces

1.1 The Concept of Continuity

The concept of continuity plays an important role in mathematics. There is a precise definition of continuity for functions of a real variable. A function f: D → ℝ, defined on a subset D of the real line ℝ, is said to be continuous at a point p of D if, given any real number ε satisfying ε > 0, there exists a real number δ satisfying δ > 0 such that |f(x) − f(p)| < ε for all points x of D satisfying |x − p| < δ. This definition of continuity can easily be adapted so as to apply to functions of a complex variable. It can also be generalized to as to apply to functions of several real or complex variables. We thus obtain a definition of continuity for functions between subsets of Euclidean spaces.

This definition of continuity generalizes directly to functions between metric spaces. A metric space is a set provided with a distance function, measuring the distance between any two points of the set. This distance function is required to satisfy certain axioms: the distance between any two points of a metric space is always non-negative, and is zero if and only if those points coincide; the distance from a point x to a point y is the same as the distance from y to x; given any three points x, y and z of a metric space, the distance from x to z is required to be less than or equal to the sum of the distance from x to y and the distance from y to z. A function from a metric space X to a metric space Y is continuous at a point p of X if and only if, given any real number ε satisfying ε > 0, there exists a real number δ satisfying δ > 0 such that the distance from f(x) to f(p) is less than ε for all points x of X whose distance from p is less than δ.

We shall introduce the concept of a topological space, and give a definition of continuity for functions from one topological space to another which generalizes the definitions of continuity discussed above for functions of a real variable, for functions of a complex variable, for functions between subsets of Euclidean spaces, and for functions from one metric space to another.

The theory of topological spaces has proved itself to be very useful in many areas of mathematics.

1.2 Topological Spaces

Definition A topological space X consists of a set X together with a collection of subsets, referred to as open sets, such that the following conditions are satisfied:—

(i) the empty set ∅ and the whole set X are open sets, (ii) the union of any collection of open sets is itself an open set,

(iii) the intersection of any finite collection of open sets is itself an open set.

The collection consisting of all the open sets in a topological space X is referred to as a topology on the set X.

Remark If it is necessary to specify explicitly the topology on a topological space then one denotes by (X, τ) the topological space whose underlying set is X and whose topology is τ. However if no confusion will arise then it is customary to denote this topological space simply by X.

1.3 Subsets of Euclidean Space

Let X be a subset of n-dimensional Euclidean space ℝ^{n}. The Euclidean distance |x − y| between two points x and y of X is defined as follows:

|x − y| = \sqrt{∑_{i=1}^{n}(x_{i} − y_{i})^{2}},

where x = (x_{1}, x_{2}, …, x_{n}) and y = (y_{1}, y_{2}, …, y_{n}). The Euclidean distances between any three points x, y and z of X satisfy the Triangle Inequality:

|x − z| ≤ |x − y| + |y − z|.

A subset V of X is said to be open in X if, given any point v of V, there exists some δ > 0 such that

{x ∈ X : |x − v| < δ} ⊂ V.

The empty set is also considered to be open in X.

Both ∅ and X are open sets in X. Also it is not difficult to show that any union of open sets in X is open in X, and that any finite intersection of open sets in X is open in X. (This will be proved in more generality for open sets in metric spaces.) Thus the collection of open sets in a subset X of a Euclidean space ℝ^{n} satisfies the topological space axioms. Thus every subset X of ℝ^{n} is a topological space with these open sets. This topology on a subset X of ℝ^{n} is referred to as the usual topology on X, generated by the Euclidean distance function.

In particular ℝ^{n} is itself a topological space.

1.4 Open Sets in Metric Spaces

Definition A metric space (X, d) consists of a set X together with a distance function d: X × X → [0, +∞) on X satisfying the following axioms:

(i) d(x, y) ≥ 0 for all x, y ∈ X,

(ii) d(x, y) = d(y, x) for all x, y ∈ X,

(iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X,

(iv) d(x, y) = 0 if and only if x = y.

The quantity d(x, y) should be thought of as measuring the distance between the points x and y. The inequality d(x, z) ≤ d(x, y)+d(y, z) is referred to as the Triangle Inequality. The elements of a metric space are usually referred to as points of that metric space.

An n-dimensional Euclidean space ℝ^{n} is a metric space with with respect to the Euclidean distance function d, defined by

d(x, y) = |x − y| = \sqrt{∑_{i=1}^{n}(x_{i} − y_{i})^{2}}

for all x, y ∈ ℝ^{n}. Any subset X of ℝ^{n} may be regarded as a metric space whose distance function is the restriction to X of the Euclidean distance function on ℝ^{n} defined above.

Definition Let (X, d) be a metric space. Given a point x of X and r ≥ 0, the open ball B_{X}(x, r) of radius r about x in X is defined by

B_{X}(x, r) ≡ {x′ ∈ X : d(x′, x) < r}.

Definition Let (X, d) be a metric space. A subset V of X is said to be an open set if and only if the following condition is satisfied:

• given any point v of V there exists some δ > 0 such that B_{X}(v, δ) ⊂ V.

By convention, we regard the empty set ∅ as being an open subset of X. (The criterion given above is satisfied vacuously in this case.)

Lemma 1.1 Let X be a metric space with distance function d, and let x_{0} be a point of X. Then, for any r > 0, the open ball B_{X}(x_{0}, r) of radius r about x_{0} is an open set in X.

Proof Let x ∈ B_{X}(x_{0}, r). We must show that there exists some δ > 0 such that B_{X}(x, δ) ⊂ B_{X}(x_{0}, r). Now d(x, x_{0}) < r, and hence δ > 0, where δ = r − d(x, x_{0}). Moreover if x′ ∈ B_{X}(x, δ) then

d(x′, x_{0}) ≤ d(x′, x) + d(x, x_{0}) < δ + d(x, x_{0}) = r,

by the Triangle Inequality, hence x′ ∈ B_{X}(x_{0}, r). Thus B_{X}(x, δ) ⊂ B_{X}(x_{0}, r), showing that B_{X}(x_{0}, r) is an open set, as required.

Proposition 1.2 Let X be a metric space. The collection of open sets in X has the following properties:—

(i) the empty set ∅ and the whole set X are both open sets;

(ii) the union of any collection of open sets is itself an open set;

(iii) the intersection of any finite collection of open sets is itself an open set.

Proof The empty set ∅ is an open set by convention. Moreover the definition of an open set is satisfied trivially by the whole set X. Thus (i) is satisfied.

Let A be any collection of open sets in X, and let U denote the union of all the open sets belonging to A. We must show that U is itself an open set. Let x ∈ U. Then x ∈ V for some open set V belonging to the collection A. Therefore there exists some δ > 0 such that B_{X}(x, δ) ⊂ V. But V ⊂ U, and thus B_{X}(x, δ) ⊂ U. This shows that U is open. Thus (ii) is satisfied.

Finally let V_{1}, V_{2}, V_{3}, …, V_{k} be a finite collection of open sets in X, and let V = V_{1} ∩ V_{2} ∩ ⋯ ∩ V_{k}. Let x ∈ V. Now x ∈ V_{j} for all j, and therefore there exist strictly positive real numbers δ_{1}, δ_{2}, …, δ_{k} such that B_{X}(x, δ_{j}) ⊂ V_{j} for j = 1, 2, …, k. Let δ be the minimum of δ_{1}, δ_{2}, …, δ_{k}. Then δ > 0. (This is where we need the fact that we are dealing with a finite collection of open sets.) Moreover B_{X}(x, δ) ⊂ B_{X}(x, δ_{j}) ⊂ V_{j} for j = 1, 2, …, k, and thus B_{X}(x, δ) ⊂ V. This shows that the intersection V of the open sets V_{1}, V_{2}, …, V_{k} is itself open. Thus (iii) is satisfied.

Any metric space may be regarded as a topological space. Indeed let X be a metric space with distance function d. We recall that a subset V of X is an open set if and only if, given any point v of V, there exists some δ > 0 such that {x ∈ X : d(x, v) < δ} ⊂ V. Proposition 1.2 shows that the topological space axioms are satisfied by the collection of open sets in any metric space. We refer to this collection of open sets as the topology generated by the distance function d on X.

1.5 Further Examples of Topological Spaces

Example Given any set X, one can define a topology on X where every subset of X is an open set. This topology is referred to as the discrete topology on X.

Example Given any set X, one can define a topology on X in which the only open sets are the empty set ∅ and the whole set X.

1.6 Closed Sets

Definition Let X be a topological space. A subset F of X is said to be a closed set if and only if its complement X \ F is an open set.

We recall that the complement of the union of some collection of subsets of some set X is the intersection of the complements of those sets, and the complement of the intersection of some collection of subsets of X is the union of the complements of those sets. The following result therefore follows directly from the definition of a topological space.

Proposition 1.3 Let X be a topological space. Then the collection of closed sets of X has the following properties:—

(i) the empty set ∅ and the whole set X are closed sets,

(ii) the intersection of any collection of closed sets is itself a closed set,

(iii) the union of any finite collection of closed sets is itself a closed set.

1.7 Hausdorff Spaces

Definition A topological space X is said to be a Hausdorff space if and only if it satisfies the following Hausdorff Axiom:

• if x and y are distinct points of X then there exist open sets U and V such that x ∈ U, y ∈ V and U ∩ V = ∅.

Lemma 1.4 All metric spaces are Hausdorff spaces.

Proof Let X be a metric space with distance function d, and let x and y be points of X, where x ≠ y. Let ε = \frac{1}{2}d(x, y). Then the open balls B_{X}(x, ε) and B_{X}(y, ε) of radius ε centred on the points x and y are open sets (see Lemma 1.1). If B_{X}(x, ε) ∩ B_{X}(y, ε) were non-empty then there would exist z ∈ X satisfying d(x, z) < ε and d(z, y) < ε. But this is impossible, since it would then follow from the Triangle Inequality that d(x, y) < 2ε, contrary to the choice of ε. Thus x ∈ B_{X}(x, ε), y ∈ B_{X}(y, ε), B_{X}(x, ε) ∩ B_{X}(y, ε) = ∅. This shows that the metric space X is a Hausdorff space.

We now give an example of a topological space which is not a Hausdorff space.

Example The Zariski topology on the set ℝ of real numbers is defined as follows: a subset U of ℝ is open (with respect to the Zariski topology) if and only if either U = ∅ or else ℝ \ U is finite. It is a straightforward exercise to verify that the topological space axioms are satisfied, so that the set ℝ of real numbers is a topological space with respect to this Zariski topology. Now the intersection of any two non-empty open sets in this topology is always non-empty. (Indeed if U and V are non-empty open sets then U = ℝ \ F_{1} and V = ℝ \ F_{2}, where F_{1} and F_{2} are finite sets of real numbers. But then U ∩ V = ℝ \ (F_{1} ∪ F_{2}), which is non-empty, since F_{1} ∪ F_{2} is finite and ℝ is infinite.) It follows immediately from this that ℝ, with the Zariski topology, is not a Hausdorff space.

1.8 Subspace Topologies

Let X be a topological space with topology τ, and let A be a subset of X. Let τ_{A} be the collection of all subsets of A that are of the form V ∩ A for V ∈ τ. Then τ_{A} is a topology on the set A. (It is a straightforward exercise to verify that the topological space axioms are satisfied.) The topology τ_{A} on A is referred to as the subspace topology on A.

Any subset of a Hausdorff space is itself a Hausdorff space (with respect to the subspace topology).

Lemma 1.5 Let X be a metric space with distance function d, and let A be a subset of X. A subset W of A is open with respect to the subspace topology on A if and only if, given any point w of W, there exists some δ > 0 such that

{a ∈ A : d(a, w) < δ} ⊂ W.

Thus the subspace topology on A coincides with the topology on A obtained on regarding A as a metric space (with respect to the distance function d).

Proof Suppose that W is open with respect to the subspace topology on A. Then there exists some open set U in X such that W = U ∩ A. Let w be a point of W. Then there exists some δ > 0 such that

{x ∈ X : d(x, w) < δ} ⊂ U.

But then

{a ∈ A : d(a, w) < δ} ⊂ U ∩ A = W.

Conversely, suppose that W is a subset of A with the property that, for any w ∈ W, there exists some δ_{w} > 0 such that

{a ∈ A : d(a, w) < δ_{w}} ⊂ W.

Define U to be the union of the open balls B_{X}(w, δ_{w}) as w ranges over all points of W, where

B_{X}(w, δ_{w}) = {x ∈ X : d(x, w) < δ_{w}}.

The set U is an open set in X, since each open ball B_{X}(w, δ_{w}) is an open set in X (Lemma 1.1), and any union of open sets is itself an open set. Moreover

B_{X}(w, δ_{w}) ∩ A = {a ∈ A : d(a, w) < δ_{w}} ⊂ W

for any w ∈ W. Therefore U ∩ A ⊂ W. However W ⊂ U ∩ A, since, W ⊂ A and {w} ⊂ B_{X}(w, δ_{w}) ⊂ U for any w ∈ W. Thus W = U ∩ A, where U is an open set in X. We deduce that W is open with respect to the subspace topology on A.

Example Let X be any subset of n-dimensional Euclidean space ℝ^{n}. Then the subspace topology on X coincides with the topology on X generated by the Euclidean distance function on X. We refer to this topology as the usual topology on X.

Let X be a topological space, and let A be a subset of X. One can readily verify the following:—

• a subset B of A is closed in A (relative to the subspace topology on A) if and only if B = A ∩ F for some closed subset F of X;

• if A is itself open in X then a subset B of A is open in A if and only if it is open in X;

• if A is itself closed in X then a subset B of A is closed in A if and only if it is closed in X.

1.9 Continuous Functions between Topological Spaces

Definition A function f: X → Y from a topological space X to a topological space Y is said to be continuous if f^{−1}(V) is an open set in X for every open set V in Y, where

f^{−1}(V) ≡ {x ∈ X : f(x) ∈ V }.

A continuous function from X to Y is often referred to as a map from X to Y.

Lemma 1.6 Let X, Y and Z be topological spaces, and let f: X → Y and g: Y → Z be continuous functions. Then the composition g ∘ f: X → Z of the functions f and g is continuous.

Proof Let V be an open set in Z. Then g^{−1}(V) is open in Y (since g is continuous), and hence f^{−1}(g^{−1}(V)) is open in X (since f is continuous). But f^{−1}(g^{−1}(V)) = (g ∘ f)^{−1}(V). Thus the composition function g ∘ f is continuous.

Lemma 1.7 Let X and Y be topological spaces, and let f: X → Y be a function from X to Y. The function f is continuous if and only if f^{−1}(G) is closed in X for every closed subset G of Y.

Proof If G is any subset of Y then X \ f^{−1}(G) = f^{−1}(Y \ G) (i.e., the complement of the preimage of G is the preimage of the complement of G). The result therefore follows immediately from the definitions of continuity and closed sets.

1.10 Continuous Functions between Metric Spaces

The following definition of continuity for functions between metric spaces generalizes that for functions of a real or complex variable.

Definition Let X and Y be metric spaces with distance functions d_{X} and d_{Y} respectively. A function f: X → Y from X to Y is said to be continuous at a point x of X if and only if the following criterion is satisfied:—

• given any real number ε satisfying ε > 0 there exists some δ > 0 such that d_{Y} (f(x), f(x′)) < ε for all points x′ of X satisfying d_{X}(x, x′) < δ.

The function f: X → Y is said to be continuous on X if and only if it is continuous at x for every point x of X.

This definition can be rephrased in terms of open balls: a function f: X → Y from a metric space X to a metric space Y is continuous at a point x of X if and only if, given any ε > 0, there exists some δ > 0 such that f maps B_{X}(x, δ) into B_{Y} (f(x), ε) (where B_{X}(x, δ) and B_{Y} (f(x), ε) denote the open balls of radius δ and ε about x and f(x) respectively).

Let f: X → Y be a function from a set X to a set Y. Given any subset V of Y, we denote by f^{−1}(V) the preimage of V under the map f, defined by

f^{−1}(V) = {x ∈ X : f(x) ∈ V }.

The following result shows that the definition of continuity given above for functions between metric spaces is consistent with the more general definition of continuity for functions between topological spaces.

Proposition 1.8 Let X and Y be metric spaces, and let f: X → Y be a function from X to Y. The function f is continuous if and only if f^{−1}(V) is an open set in X for every open set V of Y.

Proof Suppose that f: X → Y is continuous. Let V be an open set in Y. We must show that f^{−1}(V) is open in X. Let x be a point belonging to f^{−1}(V). We must show that there exists some δ > 0 with the property that B_{X}(x, δ) ⊂ f^{−1}(V). Now f(x) belongs to V. But V is open, hence there exists some ε > 0 with the property that B_{Y} (f(x), ε) ⊂ V. But f is continuous at x. Therefore there exists some δ > 0 such that f maps the open ball B_{X}(x, δ) into B_{Y} (f(x), ε) (see the remarks above). Thus f(x′) ∈ V for all x′ ∈ B_{X}(x, δ), showing that B_{X}(x, δ) ⊂ f^{−1}(V). We have thus shown that if f: X → Y is continuous then f^{−1}(V) is open in X for every open set V in Y.

Conversely suppose that f: X → Y has the property that f^{−1}(V) is open in X for every open set V in Y. Let x be any point of X. We must show that f is continuous at x. Let ε > 0 be given. The open ball B_{Y} (f(x), ε) is an open set in Y, by Lemma 1.1, hence f^{−1} (B_{Y} (f(x), ε)) is an open set in X which contains x. It follows that there exists some δ > 0 such that B_{X}(x, δ) ⊂ f^{−1} (B_{Y} (f(x), ε)). We have thus shown that, given any ε > 0, there exists some δ > 0 such that f maps the open ball B_{X}(x, δ) into B_{Y} (f(x), ε). We conclude that f is continuous at x, as required.

1.11 A Criterion for Continuity

We now show that, if a topological space X is the union of a finite collection of closed sets, and if a function from X to some topological space is continuous on each of these closed sets, then that function is continuous on X.

Lemma 1.9 Let X and Y be topological spaces, let f: X → Y be a function from X to Y, and let X = A_{1} ∪A_{2} ∪⋯∪A_{k}, where A_{1}, A_{2}, …, A_{k} are closed sets in X. Suppose that the restriction of f to the closed set A_{i} is continuous for i = 1, 2, …, k. Then f: X → Y is continuous.

Proof Let V be an open set in Y. We must show that f^{−1}(V) is open in X. Now the preimage of the open set V under the restriction f|A_{i} of f to A_{i} is f^{−1}(V) ∩ A_{i}. It follows from the continuity of f|A_{i} that f^{−1}(V) ∩ A_{i} is relatively open in A_{i} for each i, and hence there exist open sets U_{1}, U_{2}, …, U_{k} in X such that f^{−1}(V)∩A_{i} = U_{i}∩A_{i} for i = 1, 2, …, k. Let W_{i} = U_{i}∪(X\A_{i}) for i = 1, 2, …, k. Then W_{i} is an open set in X (as it is the union of the open sets U_{i} and X \ A_{i}), and W_{i} ∩ A_{i} = U_{i} ∩ A_{i} = f^{−1}(V) ∩ A_{i} for each i. We claim that f^{−1}(V) = W_{1} ∩ W_{2} ∩ ⋯ ∩ W_{k}.

Let W = W_{1} ∩ W_{2} ∩ ⋯ ∩ W_{k}. Then f^{−1}(V) ⊂ W, since f^{−1}(V) ⊂ W_{i} for each i. Also

W = ⋃_{i=1}^{k}(W ∩ A_{i}) ⊂ ⋃_{i=1}^{k}(W_{i} ∩ A_{i}) = ⋃_{i=1}^{k}(f^{−1}(V) ∩ A_{i}) ⊂ f^{−1}(V),

since X = A_{1} ∪A_{2} ∪⋯∪A_{k} and W_{i} ∩A_{i} = f^{−1}(V)∩A_{i} for each i. Therefore f^{−1}(V) = W. But W is open in X, since it is the intersection of a finite collection of open sets. We have thus shown that f^{−1}(V) is open in X for any open set V in Y. Thus f: X → Y is continuous, as required.

Alternative Proof A function f: X → Y is continuous if and only if f^{−1}(G) is closed in X for every closed set G in Y (Lemma 1.7). Let G be an closed set in Y. Then f^{−1}(G) ∩ A_{i} is relatively closed in A_{i} for i = 1, 2, …, k, since the restriction of f to A_{i} is continuous for each i. But A_{i} is closed in X, and therefore a subset of A_{i} is relatively closed in A_{i} if and only if it is closed in X. Therefore f^{−1}(G) ∩ A_{i} is closed in X for i = 1, 2, …, k. Now f^{−1}(G) is the union of the sets f^{−1}(G) ∩ A_{i} for i = 1, 2, …, k. It follows that f^{−1}(G), being a finite union of closed sets, is itself closed in X. It now follows from Lemma 1.7 that f: X → Y is continuous.

Example Let Y be a topological space, and let α: [0, 1] → Y and β: [0, 1] → Y be continuous functions defined on the interval [0, 1], where α(1) = β(0). Let γ: [0, 1] → Y be defined by

γ(t) = \{ \begin{matrix}α(2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ β(2t − 1) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

Now γ|[0, \frac{1}{2}] = α ∘ ρ where ρ: [0, \frac{1}{2}] → [0, 1] is the continuous function defined by ρ(t) = 2t for all t ∈ [0, \frac{1}{2}]. Thus γ|[0, \frac{1}{2}] is continuous, being a composition of two continuous functions. Similarly γ|[\frac{1}{2}, 1] is continuous. The subintervals [0, \frac{1}{2}] and [\frac{1}{2}, 1] are closed in [0, 1], and [0, 1] is the union of these two subintervals. It follows from Lemma 1.9 that γ: [0, 1] → Y is continuous.

1.12 Homeomorphisms

Definition Let X and Y be topological spaces. A function h: X → Y is said to be a homeomorphism if and only if the following conditions are satisfied:

• the function h: X → Y is both injective and surjective (so that the function h: X → Y has a well-defined inverse h^{−1}: Y → X),

• the function h: X → Y and its inverse h^{−1}: Y → X are both continuous.

Two topological spaces X and Y are said to be homeomorphic if there exists a homeomorphism h: X → Y from X to Y.

If h: X → Y is a homeomorphism between topological spaces X and Y then h induces a one-to-one correspondence between the open sets of X and the open sets of Y. Thus the topological spaces X and Y can be regarded as being identical as topological spaces.

1.13 Neighbourhoods, Closures and Interiors

Definition Let X be a topological space, and let x be a point of X. Let N be a subset of X which contains the point x. Then N is said to be a neighbourhood of the point x if and only if there exists an open set U for which x ∈ U and U ⊂ N.

One can readily verify that this definition of neighbourhoods in topological spaces is consistent with that for neighbourhoods in metric spaces.

Lemma 1.10 Let X be a topological space. A subset V of X is open in X if and only if V is a neighbourhood of each point belonging to V.

Proof It follows directly from the definition of neighbourhoods that an open set V is a neighbourhood of any point belonging to V. Conversely, suppose that V is a subset of X which is a neighbourhood of each v ∈ V. Then, given any point v of V, there exists an open set U_{v} such that v ∈ U_{v} and U_{v} ⊂ V. Thus V is an open set, since it is the union of the open sets U_{v} as v ranges over all points of V.

Definition Let X be a topological space and let A be a subset of X. The closure \overline{A} of A in X is defined to be the intersection of all of the closed subsets of X that contain A. The interior A^{0} of A in X is defined to be the union of all of the open subsets of X that are contained in A.

Let X be a topological space and let A be a subset of X. It follows directly from the definition of \overline{A} that the closure \overline{A} of A is uniquely characterized by the following two properties:

(i) the closure \overline{A} of A is a closed set containing A,

(ii) if F is any closed set containing A then F contains \overline{A}.

Similarly the interior A^{0} of A is uniquely characterized by the following two properties:

(i) the interior A^{0} of A is an open set contained in A,

(ii) if U is any open set contained in A then U is contained in A^{0}.

Moreover a point x of A belongs to the interior A^{0} of A if and only if A is a neighbourhood of x.

1.14 Product Topologies

The Cartesian product X_{1} × X_{2} × ⋯ × X_{n} of sets X_{1}, X_{2}, …, X_{n} is defined to be the set of all ordered n-tuples (x_{1}, x_{2}, …, x_{n}), where x_{i} ∈ X_{i} for i = 1, 2, …, n.

The sets ℝ^{2} and ℝ^{3} are the Cartesian products ℝ × ℝ and ℝ × ℝ × ℝ respectively.

Cartesian products of sets are employed as the domains of functions of several variables. For example, if X, Y and Z are sets, and if an element f(x, y) of Z is determined for each choice of an element x of X and an element y of Y, then we have a function f: X × Y → Z whose domain is the Cartesian product X × Y of X and Y: this function sends the ordered pair (x, y) to f(x, y) for all x ∈ X and y ∈ Y.

Definition Let X_{1}, X_{2}, …, X_{n} be topological spaces. A subset U of the Cartesian product X_{1} × X_{2} × ⋯ × X_{n} is said to be open (with respect to the product topology) if, given any point p of U, there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that {p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U.

Lemma 1.11 Let X_{1}, X_{2}, …, X_{n} be topological spaces. Then the collection of open sets in X_{1} × X_{2} × ⋯ × X_{n} is a topology on X_{1} × X_{2} × ⋯ × X_{n}.

Proof Let X = X_{1} ×X_{2} ×⋯×X_{n}. The definition of open sets ensures that the empty set and the whole set X are open in X. We must prove that any union or finite intersection of open sets in X is an open set.

Let E be a union of a collection of open sets in X and let p be a point of E. Then p ∈ D for some open set D in the collection. It follows from this that there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that

{p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ D ⊂ E.

Thus E is open in X.

Let U = U_{1}∩U_{2}∩⋯∩U_{m}, where U_{1}, U_{2}, …, U_{m} are open sets in X, and let p be a point of U. Then there exist open sets V_{ki} in X_{i} for k = 1, 2, …, m and i = 1, 2, …, n such that {p} ⊂ V_{k1} × V_{k2} × ⋯ × V_{kn} ⊂ U_{k} for k = 1, 2, …, m. Let V_{i} = V_{1i} ∩ V_{2i} ∩ ⋯ ∩ V_{mi} for i = 1, 2, …, n. Then

{p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ V_{k1} × V_{k2} × ⋯ × V_{kn} ⊂ U_{k}

for k = 1, 2, …, m, and hence {p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. It follows that U is open in X, as required.

Lemma 1.12 Let X_{1}, X_{2}, …, X_{n} and Z be topological spaces. Then a function f: X_{1} × X_{2} × ⋯ × X_{n} → Z is continuous if and only if, given any point p of X_{1} × X_{2} × ⋯ × X_{n}, and given any open set U in Z containing f(p), there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that p ∈ V_{1} × V_{2} ⋯ × V_{n} and f(V_{1} × V_{2} × ⋯ × V_{n}) ⊂ U.

Proof Let V_{i} be an open set in X_{i} for i = 1, 2, …, n, and let U be an open set in Z. Then V_{1}×V_{2}×⋯×V_{n} ⊂ f^{−1}(U) if and only if f(V_{1}×V_{2}×⋯×V_{n}) ⊂ U. It follows that f^{−1}(U) is open in the product topology on X_{1}×X_{2}×⋯×X_{n} if and only if, given any point p of X_{1} ×X_{2} ×⋯×X_{n} satisfying f(p) ∈ U, there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that f(V_{1} ×V_{2} ×⋯×V_{n}) ⊂ U. The required result now follows from the definition of continuity.

Let X_{1}, X_{2}, …, X_{n} be topological spaces, and let V_{i} be an open set in X_{i} for i = 1, 2, …, n. It follows directly from the definition of the product topology that V_{1} × V_{2} × ⋯ × V_{n} is open in X_{1} × X_{2} × ⋯ × X_{n}.

Theorem 1.13 Let X = X_{1} × X_{2} × ⋯ × X_{n}, where X_{1}, X_{2}, …, X_{n} are topological spaces and X is given the product topology, and for each i, let p_{i}: X → X_{i} denote the projection function which sends (x_{1}, x_{2}, …, x_{n}) ∈ X to x_{i}. Then the functions p_{1}, p_{2}, …, p_{n} are continuous. Moreover a function f: Z → X mapping a topological space Z into X is continuous if and only if p_{i} ∘ f: Z → X_{i} is continuous for i = 1, 2, …, n.

Proof Let V be an open set in X_{i}. Then

p_{i}^{−1}(V) = X_{1} × ⋯ × X_{i−1} × V × X_{i+1} × ⋯ × X_{n},

and therefore p_{i}^{−1}(V) is open in X. Thus p_{i}: X → X_{i} is continuous for all i.

Let f: Z → X be continuous. Then, for each i, p_{i} ∘ f: Z → X_{i} is a composition of continuous functions, and is thus itself continuous.

Conversely suppose that f: Z → X is a function with the property that p_{i} ∘ f is continuous for all i. Let U be an open set in X. We must show that f^{−1}(U) is open in Z.

Let z be a point of f^{−1}(U), and let f(z) = (u_{1}, u_{2}, …, u_{n}). Now U is open in X, and therefore there exist open sets V_{1}, V_{2}, …, V_{n} in X_{1}, X_{2}, …, X_{n} respectively such that u_{i} ∈ V_{i} for all i and V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. Let

N_{z} = f_{1}^{−1}(V_{1}) ∩ f_{2}^{−1}(V_{2}) ∩ ⋯ ∩ f_{n}^{−1}(V_{n}),

where f_{i} = p_{i} ∘ f for i = 1, 2, …, n. Now f_{i}^{−1}(V_{i}) is an open subset of Z for i = 1, 2, …, n, since V_{i} is open in X_{i} and f_{i}: Z → X_{i} is continuous. Thus N_{z}, being a finite intersection of open sets, is itself open in Z. Moreover

f(N_{z}) ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U,

so that N_{z} ⊂ f^{−1}(U). It follows that f^{−1}(U) is the union of the open sets N_{z} as z ranges over all points of f^{−1}(U). Therefore f^{−1}(U) is open in Z. This shows that f: Z → X is continuous, as required.

Proposition 1.14 The usual topology on ℝ^{n} coincides with the product topology on ℝ^{n} obtained on regarding ℝ^{n} as the Cartesian product ℝ × ℝ × ⋯ × ℝ of n copies of the real line ℝ.

Proof We must show that a subset U of ℝ^{n} is open with respect to the usual topology if and only if it is open with respect to the product topology.

Let U be a subset of ℝ^{n} that is open with respect to the usual topology, and let u ∈ U. Then there exists some δ > 0 such that B(u, δ) ⊂ U, where

B(u, δ) = {x ∈ ℝ^{n} : |x − u| < δ}.

Let I_{1}, I_{2}, …, I_{n} be the open intervals in ℝ defined by

I_{i} = \{t ∈ ℝ : u_{i} − \frac{δ}{\sqrt{n}} < t < u_{i} + \frac{δ}{\sqrt{n}}\} (i = 1, 2, …, n),

Then I_{1}, I_{2}, …, I_{n} are open sets in ℝ. Moreover

{u} ⊂ I_{1} × I_{2} × ⋯ × I_{n} ⊂ B(u, δ) ⊂ U,

since

|x − u|^{2} = ∑_{i=1}^{n}(x_{i} − u_{i})^{2} < n (\frac{δ}{\sqrt{n}})^{2} = δ^{2}

for all x ∈ I_{1} × I_{2} × ⋯ × I_{n}. This shows that any subset U of ℝ^{n} that is open with respect to the usual topology on ℝ^{n} is also open with respect to the product topology on ℝ^{n}.

Conversely suppose that U is a subset of ℝ^{n} that is open with respect to the product topology on ℝ^{n}, and let u ∈ U. Then there exist open sets V_{1}, V_{2}, …, V_{n} in ℝ containing u_{1}, u_{2}, …, u_{n} respectively such that V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. Now we can find δ_{1}, δ_{2}, …, δ_{n} such that δ_{i} > 0 and (u_{i} − δ_{i}, u_{i} + δ_{i}) ⊂ V_{i} for all i. Let δ > 0 be the minimum of δ_{1}, δ_{2}, …, …, δ_{n}. Then

B(u, δ) ⊂ V_{1} × V_{2} × ⋯ V_{n} ⊂ U,

for if x ∈ B(u, δ) then |x_{i} − u_{i}| < δ_{i} for i = 1, 2, …, n. This shows that any subset U of ℝ^{n} that is open with respect to the product topology on ℝ^{n} is also open with respect to the usual topology on ℝ^{n}.

The following result is now an immediate corollary of Proposition 1.14 and Theorem 1.13.

Corollary 1.15 Let X be a topological space and let f: X → ℝ^{n} be a function from X to ℝ^{n}. Let us write

f(x) = (f_{1}(x), f_{2}(x), …, f_{n}(x))

for all x ∈ X, where the components f_{1}, f_{2}, …, f_{n} of f are functions from X to ℝ. The function f is continuous if and only if its components f_{1}, f_{2}, …, f_{n} are all continuous.

Let f: X → ℝ and g: X → ℝ be continuous real-valued functions on some topological space X. We claim that f +g, f −g and f.g are continuous. Now it is a straightforward exercise to verify that the sum and product functions s: ℝ^{2} → ℝ and p: ℝ^{2} → ℝ defined by s(x, y) = x + y and p(x, y) = xy are continuous, and f + g = s ∘ h and f.g = p ∘ h, where h: X → ℝ^{2} is defined by h(x) = (f(x), g(x)). Moreover it follows from Corollary 1.15 that the function h is continuous, and compositions of continuous functions are continuous. Therefore f + g and f.g are continuous, as claimed. Also −g is continuous, and f − g = f + (−g), and therefore f − g is continuous. If in addition the continuous function g is non-zero everywhere on X then 1/g is continuous (since 1/g is the composition of g with the reciprocal function t ↦ 1/t), and therefore f/g is continuous.

Lemma 1.16 The Cartesian product X_{1} × X_{2} × … X_{n} of Hausdorff spaces X_{1}, X_{2}, …, X_{n} is Hausdorff.

Proof Let X = X_{1} × X_{2} × …, X_{n}, and let u and v be distinct points of X, where u = (x_{1}, x_{2}, …, x_{n}) and v = (y_{1}, y_{2}, …, y_{n}). Then x_{i} ≠ y_{i} for some integer i between 1 and n. But then there exist open sets U and V in X_{i} such that x_{i} ∈ U, y_{i} ∈ V and U ∩ V = ∅ (since X_{i} is a Hausdorff space). Let p_{i}: X → X_{i} denote the projection function. Then p_{i}^{−1}(U) and p_{i}^{−1}(V) are open sets in X, since p_{i} is continuous. Moreover u ∈ p_{i}^{−1}(U), v ∈ p_{i}^{−1}(V), and p_{i}^{−1}(U) ∩ p_{i}^{−1}(V) = ∅. Thus X is Hausdorff, as required.

1.15 Identification Maps and Quotient Topologies

Definition Let X and Y be topological spaces and let q: X → Y be a function from X to Y. The function q is said to be an identification map if and only if the following conditions are satisfied:

• the function q: X → Y is surjective,

• a subset U of Y is open in Y if and only if q^{−1}(U) is open in X.

It follows directly from the definition that any identification map is continuous. Moreover, in order to show that a continuous surjection q: X → Y is an identification map, it suffices to prove that if V is a subset of Y with the property that q^{−1}(V) is open in X then V is open in Y.

Lemma 1.17 Let X be a topological space, let Y be a set, and let q: X → Y be a surjection. Then there is a unique topology on Y for which the function q: X → Y is an identification map.

Proof Let τ be the collection consisting of all subsets U of Y for which q^{−1}(U) is open in X. Now q^{−1}(∅) = ∅, and q^{−1}(Y) = X, so that ∅ ∈ τ and Y ∈ τ. If {V_{α} : α ∈ A} is any collection of subsets of Y indexed by a set A, then it is a straightforward exercise to verify that

⋃_{α∈A} q^{−1}(V_{α}) = q^{−1} (⋃_{α∈A} V_{α}) , ⋂_{α∈A} q^{−1}(V_{α}) = q^{−1} (⋂_{α∈A} V_{α})

(i.e., given any collection of subsets of Y, the union of the preimages of the sets is the preimage of the union of those sets, and the intersection of the preimages of the sets is the preimage of the intersection of those sets). It follows easily from this that unions and finite intersections of sets belonging to τ must themselves belong to τ. Thus τ is a topology on Y, and the function q: X → Y is an identification map with respect to the topology τ. Clearly τ is the unique topology on Y for which the function q: X → Y is an identification map.

Let X be a topological space, let Y be a set, and let q: X → Y be a surjection. The unique topology on Y for which the function q is an identification map is referred to as the quotient topology (or identification topology) on Y.

Lemma 1.18 Let X and Y be topological spaces and let q: X → Y be an identification map. Let Z be a topological space, and let f: Y → Z be a function from Y to Z. Then the function f is continuous if and only if the composition function f ∘ q: X → Z is continuous.

Proof Suppose that f is continuous. Then the composition function f ∘ q is a composition of continuous functions and hence is itself continuous.

Conversely suppose that f ∘ q is continuous. Let U be an open set in Z. Then q^{−1}(f^{−1}(U)) is open in X (since f ∘ q is continuous), and hence f^{−1}(U) is open in Y (since the function q is an identification map). Therefore the function f is continuous, as required.

Example Let S^{1} be the unit circle in ℝ^{2}, and let q: [0, 1] → S^{1} be the map that sends t ∈ [0, 1] to (cos 2πt, sin 2πt). Then q: [0, 1] → S^{1} is an identification map, and therefore a function f: S^{1} → Z from S^{1} to some topological space Z is continuous if and only if f ∘q: [0, 1] → Z is continuous.

Example Let S^{n} be the n-sphere, consisting of all points x in ℝ^{n+1} satisfying |x| = 1. Let ℝP^{n} be the set of all lines in ℝ^{n+1} passing through the origin (i.e., ℝP^{n} is the set of all one-dimensional vector subspaces of ℝ^{n+1}). Let q: S^{n} → ℝP^{n} denote the function which sends a point x of S^{n} to the element of ℝP^{n} represented by the line in ℝ^{n+1} that passes through both x and the origin. Note that each element of ℝP^{n} is the image (under q) of exactly two antipodal points x and −x of S^{n}. The function q induces a corresponding quotient topology on ℝP^{n} such that q: S^{n} → ℝP^{n} is an identification map. The set ℝP^{n}, with this topology, is referred to as real projective nspace. In particular ℝP^{2} is referred to as the real projective plane. It follows from Lemma 1.18 that a function f: ℝP^{n} → Z from ℝP^{n} to any topological space Z is continuous if and only if the composition function f ∘ q: S^{n} → Z is continuous.

1.16 Compact Topological Spaces

Let X be a topological space, and let A be a subset of X. A collection of subsets of X in X is said to cover A if and only if every point of A belongs to at least one of these subsets. In particular, an open cover of X is collection of open sets in X that covers X.

If U and V are open covers of some topological space X then V is said to be a subcover of U if and only if every open set belonging to V also belongs to U.

Definition A topological space X is said to be compact if and only if every open cover of X possesses a finite subcover.

Lemma 1.19 Let X be a topological space. A subset A of X is compact (with respect to the subspace topology on A) if and only if, given any collection U of open sets in X covering A, there exists a finite collection V_{1}, V_{2}, …, V_{r} of open sets belonging to U such that A ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{r}.

Proof A subset B of A is open in A (with respect to the subspace topology on A) if and only if B = A ∩ V for some open set V in X. The desired result therefore follows directly from the definition of compactness.

We now show that any closed bounded interval in the real line is compact. This result is known as the Heine-Borel Theorem. The proof of this theorem uses the least upper bound principle which states that, given any non-empty set S of real numbers which is bounded above, there exists a least upper bound (or supremum) sup S for the set S.

Theorem 1.20 (Heine-Borel) Let a and b be real numbers satisfying a < b. Then the closed bounded interval [a, b] is a compact subset of ℝ.

Proof Let U be a collection of open sets in ℝ with the property that each point of the interval [a, b] belongs to at least one of these open sets. We must show that [a, b] is covered by finitely many of these open sets.

Let S be the set of all τ ∈ [a, b] with the property that [a, τ] is covered by some finite collection of open sets belonging to U, and let s = sup S. Now s ∈ W for some open set W belonging to U. Moreover W is open in ℝ, and therefore there exists some δ > 0 such that (s − δ, s + δ) ⊂ W. Moreover s − δ is not an upper bound for the set S, hence there exists some τ ∈ S satisfying τ > s − δ. It follows from the definition of S that [a, τ] is covered by some finite collection V_{1}, V_{2}, …, V_{r} of open sets belonging to U.

Let t ∈ [a, b] satisfy τ ≤ t < s + δ. Then

[a, t] ⊂ [a, τ] ∪ (s − δ, s + δ) ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{r} ∪ W,

and thus t ∈ S. In particular s ∈ S, and moreover s = b, since otherwise s would not be an upper bound of the set S. Thus b ∈ S, and therefore [a, b] is covered by a finite collection of open sets belonging to U, as required.

Lemma 1.21 Let A be a closed subset of some compact topological space X. Then A is compact.

Proof Let U be any collection of open sets in X covering A. On adjoining the open set X \ A to U, we obtain an open cover of X. This open cover of X possesses a finite subcover, since X is compact. Moreover A is covered by the open sets in the collection U that belong to this finite subcover. It follows from Lemma 1.19 that A is compact, as required.

Lemma 1.22 Let f: X → Y be a continuous function between topological spaces X and Y, and let A be a compact subset of X. Then f(A) is a compact subset of Y.

Proof Let V be a collection of open sets in Y which covers f(A). Then A is covered by the collection of all open sets of the form f^{−1}(V) for some V ∈ V. It follows from the compactness of A that there exists a finite collection V_{1}, V_{2}, …, V_{k} of open sets belonging to V such that

A ⊂ f^{−1}(V_{1}) ∪ f^{−1}(V_{2}) ∪ ⋯ ∪ f^{−1}(V_{k}).

But then f(A) ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{k}. This shows that f(A) is compact.

Lemma 1.23 Let f: X → ℝ be a continuous real-valued function on a compact topological space X. Then f is bounded above and below on X.

Proof The range f(X) of the function f is covered by some finite collection I_{1}, I_{2}, …, I_{k} of open intervals of the form (−m, m), where m ∈ ℕ, since f(X) is compact (Lemma 1.22) and ℝ is covered by the collection of all intervals of this form. It follows that f(X) ⊂ (−M, M), where (−M, M) is the largest of the intervals I_{1}, I_{2}, …, I_{k}. Thus the function f is bounded above and below on X, as required.

Proposition 1.24 Let f: X → ℝ be a continuous real-valued function on a compact topological space X. Then there exist points u and v of X such that f(u) ≤ f(x) ≤ f(v) for all x ∈ X.

Proof Let m = inf{f(x) : x ∈ X} and M = sup{f(x) : x ∈ X}. There must exist v ∈ X satisfying f(v) = M, for if f(x) < M for all x ∈ X then the function x ↦ 1/(M − f(x)) would be a continuous real-valued function on X that was not bounded above, contradicting Lemma 1.23. Similarly there must exist u ∈ X satisfying f(u) = m, since otherwise the function x ↦ 1/(f(x)−m) would be a continuous function on X that was not bounded above, again contradicting Lemma 1.23. But then f(u) ≤ f(x) ≤ f(v) for all x ∈ X, as required.

Proposition 1.25 Let A be a compact subset of a metric space X. Then A is closed in X.

Proof Let p be a point of X that does not belong to A, and let f(x) = d(x, p), where d is the distance function on X. It follows from Proposition 1.24 that there is a point q of A such that f(a) ≥ f(q) for all a ∈ A, since A is compact. Now f(q) > 0, since q ≠ p. Let δ satisfy 0 < δ ≤ f(q). Then the open ball of radius δ about the point p is contained in the complement of A, since f(x) < f(q) for all points x of this open ball. It follows that the complement of A is an open set in X, and thus A itself is closed in X.

Proposition 1.26 Let X be a Hausdorff topological space, and let K be a compact subset of X. Let x be a point of X \ K. Then there exist open sets V and W in X such that x ∈ V, K ⊂ W and V ∩ W = ∅.

Proof For each point y ∈ K there exist open sets V_{x,y} and W_{x,y} such that x ∈ V_{x,y}, y ∈ W_{x,y} and V_{x,y} ∩ W_{x,y} = ∅ (since X is a Hausdorff space). But then there exists a finite set {y_{1}, y_{2}, …, y_{r}} of points of K such that K is contained in W_{x,y_{1}} ∪ W_{x,y_{2}} ∪ ⋯ ∪ W_{x,y_{r}}, since K is compact. Define

V = V_{x,y_{1}} ∩ V_{x,y_{2}} ∩ ⋯ ∩ V_{x,y_{r}}, W = W_{x,y_{1}} ∪ W_{x,y_{2}} ∪ ⋯ ∪ W_{x,y_{r}}.

Then V and W are open sets, x ∈ V, K ⊂ W and V ∩W = ∅, as required.

Corollary 1.27 A compact subset of a Hausdorff topological space is closed.

Proof Let K be a compact subset of a Hausdorff topological space X. It follows immediately from Proposition 1.26 that, for each x ∈ X \ K, there exists an open set V_{x} such that x ∈ V_{x} and V_{x} ∩ K = ∅. But then X \ K is equal to the union of the open sets V_{x} as x ranges over all points of X \ K, and any set that is a union of open sets is itself an open set. We conclude that X \ K is open, and thus K is closed.

Proposition 1.28 Let X be a Hausdorff topological space, and let K_{1} and K_{2} be compact subsets of X, where K_{1} ∩ K_{2} = ∅. Then there exist open sets U_{1} and U_{2} such that K_{1} ⊂ U_{1}, K_{2} ⊂ U_{2} and U_{1} ∩ U_{2} = ∅.

Proof It follows from Proposition 1.26 that, for each point x of K_{1}, there exist open sets V_{x} and W_{x} such that x ∈ V_{x}, K_{2} ⊂ W_{x} and V_{x} ∩ W_{x} = ∅. But then there exists a finite set {x_{1}, x_{2}, …, x_{r}} of points of K_{1} such that

K_{1} ⊂ V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}},

since K_{1} is compact. Define

U_{1} = V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}}, U_{2} = W_{x_{1}} ∩ W_{x_{2}} ∩ ⋯ ∩ W_{x_{r}}.

Then U_{1} and U_{2} are open sets, K_{1} ⊂ U_{1}, K_{2} ⊂ U_{2} and U_{1} ∩ U_{2} = ∅, as required.

Lemma 1.29 Let f: X → Y be a continuous function from a compact topological space X to a Hausdorff space Y. Then f(K) is closed in Y for every closed set K in X.

Proof If K is a closed set in X, then K is compact (Lemma 1.21), and therefore f(K) is compact (Lemma 1.22). But any compact subset of a Hausdorff space is closed (Corollary 1.27). Thus f(K) is closed in Y, as required.

Remark If the Hausdorff space Y in Lemma 1.29 is a metric space, then Proposition 1.25 may be used in place of Corollary 1.27 in the proof of the lemma.

Theorem 1.30 A continuous bijection f: X → Y from a compact topological space X to a Hausdorff space Y is a homeomorphism.

Proof Let g: Y → X be the inverse of the bijection f: X → Y. If U is open in X then X \ U is closed in X, and hence f(X \ U) is closed in Y, by Lemma 1.29. But f(X \ U) = g^{−1}(X \ U) = Y \ g^{−1}(U). It follows that g^{−1}(U) is open in Y for every open set U in X. Therefore g: Y → X is continuous, and thus f: X → Y is a homeomorphism.

We recall that a function f: X → Y from a topological space X to a topological space Y is said to be an identification map if it is surjective and satisfies the following condition: a subset U of Y is open in Y if and only if f^{−1}(U) is open in X.

Proposition 1.31 A continuous surjection f: X → Y from a compact topological space X to a Hausdorff space Y is an identification map.

Proof Let U be a subset of Y. We claim that Y \ U = f(K), where K = X \ f^{−1}(U). Clearly f(K) ⊂ Y \ U. Also, given any y ∈ Y \ U, there exists x ∈ X satisfying y = f(x), since f: X → Y is surjective. Moreover x ∈ K, since f(x) ∉ U. Thus Y \ U ⊂ f(K), and hence Y \ U = f(K), as claimed.

We must show that the set U is open in Y if and only if f^{−1}(U) is open in X. First suppose that f^{−1}(U) is open in X. Then K is closed in X, and hence f(K) is closed in Y, by Lemma 1.29. It follows that U is open in Y. Conversely if U is open in Y then f^{−1}(U) is open in X, since f: X → Y is continuous. Thus the surjection f: X → Y is an identification map.

Example Let S^{1} be the unit circle in ℝ^{2}, defined by S^{1} = {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} = 1}, and let q: [0, 1] → S^{1} be defined by q(t) = (cos 2πt, sin 2πt) for all t ∈ [0, 1]. It has been shown that the map q is an identification map. This also follows directly from the fact that q: [0, 1] → S^{1} is a continuous surjection from the compact space [0, 1] to the Hausdorff space S^{1}.

We shall show that a finite Cartesian product of compact spaces is compact. To prove this, we apply the following result, known as the Tube Lemma.

Lemma 1.32 Let X and Y be topological spaces, let K be a compact subset of Y, and U be an open set in X × Y. Let V = {x ∈ X : {x} × K ⊂ U}. Then V is an open set in X.

Proof Let x ∈ V. For each y ∈ K there exist open subsets D_{y} and E_{y} of X and Y respectively such that (x, y) ∈ D_{y} × E_{y} and D_{y} × E_{y} ⊂ U. Now there exists a finite set {y_{1}, y_{2}, …, y_{k}} of points of K such that K ⊂ E_{y_{1}} ∪ E_{y_{2}} ∪ ⋯ ∪ E_{y_{k}}, since K is compact. Set N_{x} = D_{y_{1}} ∩ D_{y_{2}} ∩ ⋯ ∩ D_{y_{k}}. Then N_{x} is an open set in X. Moreover

N_{x} × K ⊂ ⋃_{i=1}^{k}(N_{x} × E_{y_{i}}) ⊂ ⋃_{i=1}^{k}(D_{y_{i}} × E_{y_{i}}) ⊂ U,

so that N_{x} ⊂ V. It follows that V is the union of the open sets N_{x} for all x ∈ V. Thus V is itself an open set in X, as required.

Theorem 1.33 A Cartesian product of a finite number of compact spaces is itself compact.

Proof It suffices to prove that the product of two compact topological spaces X and Y is compact, since the general result then follows easily by induction on the number of compact spaces in the product.

Let U be an open cover of X × Y. We must show that this open cover possesses a finite subcover.

Let x be a point of X. The set {x}×Y is a compact subset of X ×Y, since it is the image of the compact space Y under the continuous map from Y to X × Y which sends y ∈ Y to (x, y), and the image of any compact set under a continuous map is itself compact (Lemma 1.22). Therefore there exists a finite collection U_{1}, U_{2}, …, U_{r} of open sets belonging to the open cover U such that {x} × Y is contained in U_{1} ∪ U_{2} ∪ ⋯ ∪ U_{r}. Let V_{x} denote the set of all points x′ of X for which {x′} × Y is contained in U_{1} ∪ U_{2} ∪ ⋯ ∪ U_{r}. Then x ∈ V_{x}, and Lemma 1.32 ensures That V_{x} is an open set in X. Note that V_{x} × Y is covered by finitely many of the open sets belonging to the open cover U.

Now {V_{x} : x ∈ X} is an open cover of the space X. It follows from the compactness of X that there exists a finite set {x_{1}, x_{2}, …, x_{r}} of points of X such that X = V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}}. Now X × Y is the union of the sets V_{x_{j}} × Y for j = 1, 2, …, r, and each of these sets can be covered by a finite collection of open sets belonging to the open cover U. On combining these finite collections, we obtain a finite collection of open sets belonging to U which covers X × Y. This shows that X × Y is compact.

Theorem 1.34 Let K be a subset of ℝ^{n}. Then K is compact if and only if K is both closed and bounded.

Proof Suppose that K is compact. Then K is closed, since ℝ^{n} is Hausdorff, and a compact subset of a Hausdorff space is closed (by Corollary 1.27). For each natural number m, let B_{m} be the open ball of radius m about the origin, given by B_{m} = {x ∈ ℝ^{n} : |x| < m}. Then {B_{m} : m ∈ ℕ} is an open cover of ℝ^{n}. It follows from the compactness of K that there exist natural numbers m_{1}, m_{2}, …, m_{k} such that K ⊂ B_{m_{1}} ∪ B_{m_{2}} ∪ ⋯ ∪ B_{m_{k}}. But then K ⊂ B_{M}, where M is the maximum of m_{1}, m_{2}, …, m_{k}, and thus K is bounded.

Conversely suppose that K is both closed and bounded. Then there exists some real number L such that K is contained within the closed cube C given by

C = {(x_{1}, x_{2}, …, x_{n}) ∈ ℝ^{n} : −L ≤ x_{j} ≤ L for j = 1, 2, …, n}.

Now the closed interval [−L, L] is compact by the Heine-Borel Theorem (Theorem 1.20), and C is the Cartesian product of n copies of the compact set [−L, L]. It follows from Theorem 1.33 that C is compact. But K is a closed subset of C, and a closed subset of a compact topological space is itself compact, by Lemma 1.21. Thus K is compact, as required.

1.17 The Lebesgue Lemma and Uniform Continuity

Definition Let X be a metric space with distance function d. A subset A of X is said to be bounded if there exists a non-negative real number K such that d(x, y) ≤ K for all x, y ∈ A. The smallest real number K with this property is referred to as the diameter of A, and is denoted by diam A. (Note that diam A is the supremum of the values of d(x, y) as x and y range over all points of A.)

Lemma 1.35 (Lebesgue Lemma) Let (X, d) be a compact metric space. Let U be an open cover of X. Then there exists a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U.

Proof Every point of X is contained in at least one of the open sets belonging to the open cover U. It follows from this that, for each point x of X, there exists some δ_{x} > 0 such that the open ball B(x, 2δ_{x}) of radius 2δ_{x} about the point x is contained wholly within one of the open sets belonging to the open cover U. But then the collection consisting of the open balls B(x, δ_{x}) of radius δ_{x} about the points x of X forms an open cover of the compact space X. Therefore there exists a finite set x_{1}, x_{2}, …, x_{r} of points of X such that

B(x_{1}, δ_{1}) ∪ B(x_{2}, δ_{2}) ∪ ⋯ ∪ B(x_{r}, δ_{r}) = X,

where δ_{i} = δ_{x_{i}} for i = 1, 2, …, r. Let δ > 0 be given by

δ = minimum(δ_{1}, δ_{2}, …, δ_{r}).

Suppose that A is a subset of X whose diameter is less than δ. Let u be a point of A. Then u belongs to B(x_{i}, δ_{i}) for some integer i between 1 and r. But then it follows that A ⊂ B(x_{i}, 2δ_{i}), since, for each point v of A,

d(v, x_{i}) ≤ d(v, u) + d(u, x_{i}) < δ + δ_{i} ≤ 2δ_{i}.

But B(x_{i}, 2δ_{i}) is contained wholly within one of the open sets belonging to the open cover U. Thus A is contained wholly within one of the open sets belonging to U, as required.

Let U be an open cover of a compact metric space X. A Lebesgue number for the open cover U is a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U. The Lebesgue Lemma thus states that there exists a Lebesgue number for every open cover of a compact metric space.

Let X and Y be metric spaces with distance functions d_{X} and d_{Y} respectively, and let f: X → Y be a function from X to Y. The function f is said to be uniformly continuous on X if and only if, given ε > 0, there exists some δ > 0 such that d_{Y} (f(x), f(x′)) < ε for all points x and x′ of X satisfying d_{X}(x, x′) < δ. (The value of δ should be independent of both x and x′.)

Theorem 1.36 Let X and Y be metric spaces. Suppose that X is compact. Then every continuous function from X to Y is uniformly continuous.

Proof Let d_{X} and d_{Y} denote the distance functions for the metric spaces X and Y respectively. Let f: X → Y be a continuous function from X to Y. We must show that f is uniformly continuous.

Let ε > 0 be given. For each y ∈ Y, define

V_{y} = \{x ∈ X : d_{Y} (f(x), y) < \frac{1}{2}ε\}.

Note that V_{y} = f^{−1} (B_{Y} (y, \frac{1}{2}ε)), where B_{Y} (y, \frac{1}{2}ε) denotes the open ball of radius \frac{1}{2}ε about y in Y. Now the open ball B_{Y} (y, \frac{1}{2}ε) is an open set in Y, and f is continuous. Therefore V_{y} is open in X for all y ∈ Y. Note that x ∈ V_{f(x)} for all x ∈ X.

Now {V_{y} : y ∈ Y } is an open cover of the compact metric space X. It follows from the Lebesgue Lemma (Lemma 1.35) that there exists some δ > 0 such that every subset of X whose diameter is less than δ is a subset of some set V_{y}. Let x and x′ be points of X satisfying d_{X}(x, x′) < δ. The diameter of the set {x, x′} is d_{X}(x, x′), which is less than δ. Therefore there exists some y ∈ Y such that x ∈ V_{y} and x′ ∈ V_{y}. But then d_{Y} (f(x), y) < \frac{1}{2}ε and d_{Y} (f(x′), y) < \frac{1}{2}ε, and hence

d_{Y} (f(x), f(x′)) ≤ d_{Y} (f(x), y) + d_{Y} (y, f(x′)) < ε.

This shows that f: X → Y is uniformly continuous, as required.

Let K be a closed bounded subset of ℝ^{n}. It follows from Theorem 1.34) and Theorem 1.36 that any continuous function f: K → ℝ^{k} is uniformly continuous.

1.18 Connected Topological Spaces

Definition A topological space X is said to be connected if the empty set ∅ and the whole space X are the only subsets of X that are both open and closed.

Lemma 1.37 A topological space X is connected if and only if it has the following property: if U and V are non-empty open sets in X such that X = U ∪ V, then U ∩ V is non-empty.

Proof If U is a subset of X that is both open and closed, and if V = X \ U, then U and V are both open, U ∪ V = X and U ∩ V = ∅. Conversely if U and V are open subsets of X satisfying U ∪ V = X and U ∩ V = ∅, then U = X \V, and hence U is both open and closed. Thus a topological space X is connected if and only if there do not exist non-empty open sets U and V such that U ∪ V = X and U ∩ V = ∅. The result follows.

Let ℤ be the set of integers with the usual topology (i.e., the subspace topology on ℤ induced by the usual topology on ℝ). Then {n} is open for all n ∈ ℤ, since

\{n\} = ℤ ∩ \{t ∈ ℝ : |t − n| < \frac{1}{2}\}.

It follows that every subset of ℤ is open (since it is a union of sets consisting of a single element, and any union of open sets is open). It follows that a function f: X → ℤ on a topological space X is continuous if and only if f^{−1}(V) is open in X for any subset V of ℤ. We use this fact in the proof of the next theorem.

Proposition 1.38 A topological space X is connected if and only if every continuous function f: X → ℤ from X to the set ℤ of integers is constant.

Proof Suppose that X is connected. Let f: X → ℤ be a continuous function. Choose n ∈ f(X), and let

U = {x ∈ X : f(x) = n}, V = {x ∈ X : f(x) ≠ n}.

Then U and V are the preimages of the open subsets {n} and ℤ \ {n} of ℤ, and therefore both U and V are open in X. Moreover U ∩ V = ∅, and X = U ∪ V. It follows that V = X \ U, and thus U is both open and closed. Moreover U is non-empty, since n ∈ f(X). It follows from the connectedness of X that U = X, so that f: X → ℤ is constant, with value n.

Conversely suppose that every continuous function f: X → ℤ is constant. Let S be a subset of X which is both open and closed. Let f: X → ℤ be defined by

f(x) = \{ \begin{matrix}1 & if x ∈ S; \\ 0 & if x ∉ S. \end{matrix}

Now the preimage of any subset of ℤ under f is one of the open sets ∅, S, X \ S and X. Therefore the function f is continuous. But then the function f is constant, so that either S = ∅ or S = X. This shows that X is connected.

Lemma 1.39 The closed interval [a, b] is connected, for all real numbers a and b satisfying a ≤ b.

Proof Let f: [a, b] → ℤ be a continuous integer-valued function on [a, b]. We show that f is constant on [a, b]. Indeed suppose that f were not constant. Then f(τ) ≠ f(a) for some τ ∈ [a, b]. But the Intermediate Value Theorem would then ensure that, given any real number c between f(a) and f(τ), there would exist some t ∈ [a, τ] for which f(t) = c, and this is clearly impossible, since f is integer-valued. Thus f must be constant on [a, b]. We now deduce from Proposition 1.38 that [a, b] is connected.

Example Let X = {(x, y) ∈ ℝ^{2} : x ≠ 0}. The topological space X is not connected. Indeed if f: X → ℤ is defined by

f(x, y) = \{ \begin{matrix}1 & if x > 0, \\ −1 & if x < 0, \end{matrix}

then f is continuous on X but is not constant.

A concept closely related to that of connectedness is path-connectedness. Let x_{0} and x_{1} be points in a topological space X. A path in X from x_{0} to x_{1} is defined to be a continuous function γ: [0, 1] → X such that γ(0) = x_{0} and γ(1) = x_{1}. A topological space X is said to be path-connected if and only if, given any two points x_{0} and x_{1} of X, there exists a path in X from x_{0} to x_{1}.

Proposition 1.40 Every path-connected topological space is connected.

Proof Let X be a path-connected topological space, and let f: X → ℤ be a continuous integer-valued function on X. If x_{0} and x_{1} are any two points of X then there exists a path γ: [0, 1] → X such that γ(0) = x_{0} and γ(1) = x_{1}. But then f ∘ γ: [0, 1] → ℤ is a continuous integer-valued function on [0, 1]. But [0, 1] is connected (Lemma 1.39), therefore f∘γ is constant (Proposition 1.38). It follows that f(x_{0}) = f(x_{1}). Thus every continuous integer-valued function on X is constant. Therefore X is connected, by Proposition 1.38.

The topological spaces ℝ, ℂ and ℝ^{n} are all path-connected. Indeed, given any two points of one of these spaces, the straight line segment joining these two points is a continuous path from one point to the other. Also the n-sphere S^{n} is path-connected for all n > 0. We conclude that these topological spaces are connected.

Let A be a subset of a topological space X. Using Lemma 1.37 and the definition of the subspace topology, we see that A is connected if and only if the following condition is satisfied:

• if U and V are open sets in X such that A∩U and A∩V are non-empty and A ⊂ U ∪ V then A ∩ U ∩ V is also non-empty.

Lemma 1.41 Let X be a topological space and let A be a connected subset of X. Then the closure \overline{A} of A is connected.

Proof It follows from the definition of the closure of A that \overline{A} ⊂ F for any closed subset F of X for which A ⊂ F. On taking F to be the complement of some open set U, we deduce that \overline{A} ∩ U = ∅ for any open set U for which A ∩ U = ∅. Thus if U is an open set in X and if \overline{A} ∩ U is non-empty then A ∩ U must also be non-empty.

Now let U and V be open sets in X such that \overline{A} ∩ U and \overline{A} ∩ V are non-empty and \overline{A} ⊂ U ∪ V. Then A ∩ U and A ∩ V are non-empty, and A ⊂ U ∪ V. But A is connected. Therefore A ∩ U ∩ V is non-empty, and thus \overline{A} ∩ U ∩ V is non-empty. This shows that \overline{A} is connected.

Lemma 1.42 Let f: X → Y be a continuous function between topological spaces X and Y, and let A be a connected subset of X. Then f(A) is connected.

Proof Let g: f(A) → ℤ be any continuous integer-valued function on f(A). Then g ∘ f: A → ℤ is a continuous integer-valued function on A. It follows from Proposition 1.38 that g ∘ f is constant on A. Therefore g is constant on f(A). We deduce from Proposition 1.38 that f(A) is connected.

Lemma 1.43 The Cartesian product X × Y of connected topological spaces X and Y is itself connected.

Proof Let f: X×Y → ℤ be a continuous integer-valued function from X×Y to Z. Choose x_{0} ∈ X and y_{0} ∈ Y. The function x ↦ f(x, y_{0}) is continuous on X, and is thus constant. Therefore f(x, y_{0}) = f(x_{0}, y_{0}) for all x ∈ X. Now fix x. The function y ↦ f(x, y) is continuous on Y, and is thus constant. Therefore

f(x, y) = f(x, y_{0}) = f(x_{0}, y_{0})

for all x ∈ X and y ∈ Y. We deduce from Proposition 1.38 that X × Y is connected.

We deduce immediately that a finite Cartesian product of connected topological spaces is connected.

Proposition 1.44 Let X be a topological space. For each x ∈ X, let S_{x} be the union of all connected subsets of X that contain x. Then

(i) S_{x} is connected,

(ii) S_{x} is closed,

(iii) if x, y ∈ X, then either S_{x} = S_{y}, or else S_{x} ∩ S_{y} = ∅.

Proof Let f: S_{x} → ℤ be a continuous integer-valued function on S_{x}, for some x ∈ X. Let y be any point of S_{x}. Then, by definition of S_{x}, there exists some connected set A containing both x and y. But then f is constant on A, and thus f(x) = f(y). This shows that the function f is constant on S_{x}. We deduce that S_{x} is connected. This proves (i). Moreover the closure \overline{S_{x}} is connected, by Lemma 1.41. Therefore \overline{S_{x}} ⊂ S_{x}. This shows that S_{x} is closed, proving (ii).

Finally, suppose that x and y are points of X for which S_{x} ∩ S_{y} ≠ ∅. Let f: S_{x} ∪ S_{y} → ℤ be any continuous integer-valued function on S_{x} ∪ S_{y}. Then f is constant on both S_{x} and S_{y}. Moreover the value of f on S_{x} must agree with that on S_{y}, since S_{x} ∩ S_{y} is non-empty. We deduce that f is constant on S_{x} ∪ S_{y}. Thus S_{x} ∪ S_{y} is a connected set containing both x and y, and thus S_{x} ∪ S_{y} ⊂ S_{x} and S_{x} ∪ S_{y} ⊂ S_{y}, by definition of S_{x} and S_{y}. We conclude that S_{x} = S_{y}. This proves (iii).

Given any topological space X, the connected subsets S_{x} of X defined as in the statement of Proposition 1.44 are referred to as the connected components of X. We see from Proposition 1.44, part (iii) that the topological space X is the disjoint union of its connected components.

Example The connected components of {(x, y) ∈ ℝ^{2} : x ≠ 0} are

{(x, y) ∈ ℝ^{2} : x > 0} and {(x, y) ∈ ℝ^{2} : x < 0}.

Example The connected components of

\{t ∈ ℝ : |t − n| < \frac{1}{2} for some integer n\}.

are the sets J_{n} for all n ∈ ℤ, where J_{n} = (n − \frac{1}{2}, n + \frac{1}{2}).

2 Homotopies and Covering Maps

2.1 Homotopies

Definition Let f: X → Y and g: X → Y be continuous maps between topological spaces X and Y. The maps f and g are said to be homotopic if there exists a continuous map H: X × [0, 1] → Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x ∈ X. If the maps f and g are homotopic then we denote this fact by writing f ≃ g. The map H with the properties stated above is referred to as a homotopy between f and g.

Continuous maps f and g from X to Y are homotopic if and only if it is possible to ‘continuously deform’ the map f into the map g.

Lemma 2.1 Let X and Y be topological spaces. The homotopy relation ≃ is an equivalence relation on the set of all continuous maps from X to Y.

Proof Clearly f ≃ f, since (x, t) ↦ f(x) is a homotopy between f and itself. Thus the relation is reflexive. If f ≃ g then there exists a homotopy H: X × [0, 1] → Y between f and g (so that H(x, 0) = f(x) and H(x, 1) = g(x) for all x ∈ X). But then (x, t) ↦ H(x, 1 − t) is a homotopy between g and f. Therefore f ≃ g if and only if g ≃ f. Thus the relation is symmetric. Finally, suppose that f ≃ g and g ≃ h. Then there exist homotopies H_{1}: X × [0, 1] → Y and H_{2}: X × [0, 1] → Y such that H_{1}(x, 0) = f(x), H_{1}(x, 1) = g(x) = H_{2}(x, 0) and H_{2}(x, 1) = h(x) for all x ∈ X. Define H: X × [0, 1] → Y by

H(x, t) = \{ \begin{matrix}H_{1}(x, 2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ H_{2}(x, 2t − 1) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

Now H|X ×[0, \frac{1}{2}] and H|X ×[\frac{1}{2}, 1] are continuous. It follows from elementary point set topology that H is continuous on X × [0, 1]. Moreover H(x, 0) = f(x) and H(x, 1) = h(x) for all x ∈ X. Thus f ≃ h. Thus the relation is transitive. The relation ≃ is therefore an equivalence relation.

Definition Let X and Y be topological spaces, and let A be a subset of X. Let f: X → Y and g: X → Y be continuous maps from X to some topological space Y, where f|A = g|A (i.e., f(a) = g(a) for all a ∈ A). We say that f and g are homotopic relative to A (denoted by f ≃ g rel A) if and only if there exists a (continuous) homotopy H: X × [0, 1] → Y such that H(x, 0) = f(x) and H(x, 1) = g(x) for all x ∈ X and H(a, t) = f(a) = g(a) for all a ∈ A.

Homotopy relative to a chosen subset of X is also an equivalence relation on the set of all continuous maps between topological spaces X and Y.

2.2 Covering Maps

Definition Let X and X̃ be topological spaces and let p: X̃ → X be a continuous map. An open subset U of X is said to be evenly covered by the map p if and only if p^{−1}(U) is a disjoint union of open sets of X̃ each of which is mapped homeomorphically onto U by p. The map p: X̃ → X is said to be a covering map if p: X̃ → X is surjective and in addition every point of X is contained in some open set that is evenly covered by the map p. If p: X̃ → X is a covering map, then we say that X̃ is a covering space of X.

Example Let S^{1} be the unit circle in ℝ^{2}. Then the map p: ℝ → S^{1} defined by

p(t) = (cos 2πt, sin 2πt)

is a covering map. Indeed let n be a point of S^{1}. Consider the open set U in S^{1} containing n defined by U = S^{1} \ {−n}. Now n = (cos 2πt_{0}, sin 2πt_{0}) for some t_{0} ∈ ℝ. Then p^{−1}(U) is the union of the disjoint open sets J_{n} for all integers n, where

J_{n} = \{t ∈ ℝ : t_{0} + n − \frac{1}{2} < t < t_{0} + n + \frac{1}{2}\}.

Each of the open sets J_{n} is mapped homeomorphically onto U by the map p. This shows that p: ℝ → S^{1} is a covering map.

Example The map p: ℂ → ℂ \ {0} defined by p(z) = exp(z) is a covering map. Indeed, given any θ ∈ [−π, π] let us define

U_{θ} = {z ∈ ℂ \ {0} : arg(−z) ≠ θ}.

Then p^{−1}(U_{θ}) is the disjoint union of the open sets

{z ∈ ℂ : |Im z − θ − 2πn| < π} ,

for all integers n, and p maps each of these open sets homeomorphically onto U_{θ}. Thus U_{θ} is evenly covered by the map p.

Example Consider the map α: (−2, 2) → S^{1}, where α(t) = (cos 2πt, sin 2πt) for all t ∈ (−2, 2). It can easily be shown that there is no open set U containing the point (1, 0) that is evenly covered by the map α. Indeed suppose that there were to exist such an open set U. Then there would exist some δ satisfying 0 < δ < \frac{1}{2} such that U_{δ} ⊂ U, where

U_{δ} = {(cos 2πt, sin 2πt) : −δ < t < δ}.

The open set U_{δ} would then be evenly covered by the map α. However the connected components of α^{−1}(U_{δ}) are (−2, −2 + δ), (−1 − δ, −1 + δ), (−δ, δ), (1 − δ, 1 + δ) and (2 − δ, 2), and neither (−2, −2 + δ) nor (2 − δ, 2) is mapped homeomorphically onto U_{δ} by α.

Lemma 2.2 Let p: X̃ → X be a covering map. Then p(V) is open in X for every open set V in X̃. In particular, a covering map p: X̃ → X is a homeomorphism if and only if it is a bijection.

Proof Let V be open in X̃, and let x ∈ p(V). Then x = p(v) for some v ∈ V. Now there exists an open set U containing the point x which is evenly covered by the covering map p. Then p^{−1}(U) is a disjoint union of open sets, each of which is mapped homeomorphically onto U by the covering map p. One of these open sets contains v; let Ũ be this open set, and let N_{x} = p(V ∩ Ũ). Now N_{x} is open in X, since V ∩ Ũ is open in Ũ and p|Ũ is a homeomorphism from Ũ to U. Also x ∈ N_{x} and N_{x} ⊂ p(V). It follows that p(V) is the union of the open sets N_{x} as x ranges over all points of p(V), and thus p(V) is itself an open set, as required. The result that a bijective covering map is a homeomorphism then follows directly from the fact that a continuous bijection is a homeomorphism if and only if it maps open sets to open sets.

2.3 Path Lifting and the Monodromy Theorem

Let p: X̃ → X be a covering map over a topological space X. Let Z be a topological space, and let f: Z → X be a continuous map from Z to X. A continuous map f̃: Z → X̃ is said to be a lift of the map f: Z → X if and only if p ∘ f̃ = f. We shall prove various results concerning the existence and uniqueness of such lifts.

Proposition 2.3 Let p: X̃ → X be a covering map, let Z be a connected topological space, and let g: Z → X̃ and h: Z → X̃ be continuous maps. Suppose that p ∘ g = p ∘ h and that g(z) = h(z) for some z ∈ Z. Then g = h.

Proof Let Z_{0} = {z ∈ Z : g(z) = h(z)}. Note that Z_{0} is non-empty, by hypothesis. We show that Z_{0} is both open and closed in Z.

Let z be a point of Z. There exists an open set U in X containing the point p(g(z)) which is evenly covered by the covering map p. Then p^{−1}(U) is a disjoint union of open sets, each of which is mapped homeomorphically onto U by the covering map p. One of these open sets contains g(z); let this set be denoted by Ũ. Also one of these open sets contains h(z); let this open set be denoted by Ṽ. Let N_{z} = g^{−1}(Ũ) ∩ h^{−1}(Ṽ). Then N_{z} is an open set in Z containing z.

Consider the case when z ∈ Z_{0}. Then g(z) = h(z), and therefore Ṽ = Ũ. It follows from this that both g and h map the open set N_{z} into Ũ. But p ∘ g = p ∘ h, and p|Ũ: Ũ → U is a homeomorphism. Therefore g|N_{z} = h|N_{z}, and thus N_{z} ⊂ Z_{0}. We have thus shown that, for each z ∈ Z_{0}, there exists an open set N_{z} such that z ∈ N_{z} and N_{z} ⊂ Z_{0}. We conclude that Z_{0} is open.

Next consider the case when z ∈ Z \ Z_{0}. In this case Ũ ∩ Ṽ = ∅, since g(z) ≠ h(z). But g(N_{z}) ⊂ Ũ and h(N_{z}) ⊂ Ṽ. Therefore g(z′) ≠ h(z′) for all z′ ∈ N_{z}, and thus N_{z} ⊂ Z \Z_{0}. We have thus shown that, for each z ∈ Z \Z_{0}, there exists an open set N_{z} such that z ∈ N_{z} and N_{z} ⊂ Z \ Z_{0}. We conclude that Z \ Z_{0} is open.

The subset Z_{0} of Z is therefore both open and closed. Also Z_{0} is nonempty by hypothesis. We deduce that Z_{0} = Z, since Z is connected. Thus g = h, as required.

Lemma 2.4 Let p: X̃ → X be a covering map, let Z be a topological space, let A be a connected subset of Z, and let f: Z → X and g: A → X̃ be continuous maps with the property that p ∘ g = f|A. Suppose that f(Z) ⊂ U, where U is an open subset of X that is evenly covered by the covering map p. Then there exists a continuous map f̃: Z → X̃ such that f̃|A = g and p ∘ f̃ = f.

Proof The open set U is evenly covered by the covering map p, and therefore p^{−1}(U) is a disjoint union of open sets, each of which is mapped homeomorphically onto U by the covering map p. One of these open sets contains g(a) for some a ∈ A; let this set be denoted by Ũ. Let σ: U → Ũ be the inverse of the homeomorphism p|Ũ: Ũ → U, and let f̃ = σ ∘ f. Then p ∘ f̃ = f. Also p ∘ f̃|A = p ∘ g and f̃(a) = g(a). It follows from Proposition 2.3 that f̃|A = g, since A is connected. Thus f̃: Z → X̃ is the required map.

Theorem 2.5 (Path Lifting Theorem) Let p: X̃ → X be a covering map, let γ: [0, 1] → X be a continuous path in X, and let w be a point of X̃ satisfying p(w) = γ(0). Then there exists a unique continuous path γ̃: [0, 1] → X̃ such that γ̃(0) = w and p ∘ γ̃ = γ.

Proof The map p: X̃ → X is a covering map; therefore there exists an open cover U of X such that each open set U belonging to X is evenly covered by the map p. Now the collection consisting of the preimages γ^{−1}(U) of the open sets U belonging to U is an open cover of the interval [0, 1]. But [0, 1] is compact, by the Heine-Borel Theorem. It follows from the Lebesgue Lemma that there exists some δ > 0 such that every subinterval of length less than δ is mapped by γ into one of the open sets belonging to U. Partition the interval [0, 1] into subintervals [t_{i−1}, t_{i}], where 0 = t_{0} < t_{1} < ⋯ < t_{n−1} < t_{n} = 1, and where the length of each subinterval is less than δ. Then each subinterval [t_{i−1}, t_{i}] is mapped by γ into some open set in X that is evenly covered by the map p. It follows from Lemma 2.4 that once γ̃(t_{i−1}) has been determined, we can extend γ̃ continuously over the ith subinterval [t_{i−1}, t_{i}]. Thus by extending γ̃ successively over [t_{0}, t_{1}], [t_{1}, t_{2}],…, [t_{n−1}, t_{n}], we can lift the path γ: [0, 1] → X to a path γ̃: [0, 1] → X̃ starting at w. The uniqueness of γ̃ follows from Proposition 2.3.

Theorem 2.6 (The Monodromy Theorem) Let p: X̃ → X be a covering map, let H: [0, 1] × [0, 1] → X be a continuous map, and let w be a point of X̃ satisfying p(w) = H(0, 0). Then there exists a unique continuous map H̃: [0, 1] × [0, 1] → X̃ such that H̃(0, 0) = w and p ∘ H̃ = H.

Proof The unit square [0, 1] × [0, 1] is compact. By applying the Lebesgue Lemma to an open cover of the square by preimages of evenly covered open sets in X (as in the proof of Theorem 2.5), we see that there exists some δ > 0 with the property that any square contained in [0, 1] × [0, 1] whose sides have length less than δ is mapped by H into some open set in X which is evenly covered by the covering map p. It follows from Lemma 2.4 that if the lift H̃ of H has already been determined over a corner, or along one side, or along two adjacent sides of a square whose sides have length less than δ, then H̃ can be extended over the whole of that square. Thus if we subdivide [0, 1] × [0, 1] into squares S_{j,k}, where

S_{j,k} = \{(s, t) ∈ [0, 1] × [0, 1] : \frac{j − 1}{n} ≤ s ≤ \frac{j}{n} and \frac{k − 1}{n} ≤ t ≤ \frac{k}{n}\} ,

and 1/n < δ, then we can extend the map g to a lift H̃ of H by successively extending H̃ in turn over each of these smaller squares. (Indeed the map H̃ can be extended successively over the squares

S_{1,1}, S_{1,2}, …, S_{1,n}, S_{2,1}, S_{2,2}, …, S_{2,n}, S_{3,1}, …, S_{n−1,n}, …, S_{n,1}, S_{n,2}, …, S_{n,n}.)

The uniqueness of H̃ follows from Proposition 2.3.

3 The Fundamental Group

3.1 The Fundamental Group of a Topological Space

Definition Let X be a topological space, and let x_{0} and x_{1} be points of X. A path in X from x_{0} to x_{1} is defined to be a continuous map γ: [0, 1] → X for which γ(0) = x_{0} and γ(1) = x_{1}. A loop in X based at x_{0} is defined to be a continuous map γ: [0, 1] → X for which γ(0) = γ(1) = x_{0}.

We can concatenate paths. Let γ_{1}: [0, 1] → X and γ_{2}: [0, 1] → X be paths in some topological space X. Suppose that γ_{1}(1) = γ_{2}(0). We define the product path γ_{1}.γ_{2}: [0, 1] → X by

(γ_{1}.γ_{2})(t) = \{ \begin{matrix}γ_{1}(2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ γ_{2}(2t − 1) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

(The continuity of γ_{1}.γ_{2} may be deduced from Lemma 2.1.)

If γ: [0, 1] → X is a path in X then we define the inverse path γ^{−1}: [0, 1] → X by γ^{−1}(t) = γ(1−t). (Thus if γ is a path from the point x_{0} to the point x_{1} then γ^{−1} is the path from x_{1} to x_{0} obtained by traversing γ in the reverse direction.)

Let X be a topological space, and let x_{0} ∈ X be some chosen point of X. We define an equivalence relation on the set of all (continuous) loops based at the basepoint x_{0} of X, where two such loops γ_{0} and γ_{1} are equivalent if and only if γ_{0} ≃ γ_{1} rel {0, 1}. We denote the equivalence class of a loop γ: [0, 1] → X based at x_{0} by [γ]. This equivalence class is referred to as the based homotopy class of the loop γ. The set of equivalence classes of loops based at x_{0} is denoted by π_{1}(X, x_{0}). Thus two loops γ_{0} and γ_{1} represent the same element of π_{1}(X, x_{0}) if and only if γ_{0} ≃ γ_{1} rel {0, 1} (i.e., there exists a homotopy F: [0, 1] × [0, 1] → X between γ_{0} and γ_{1} which maps (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]).

Theorem 3.1 Let X be a topological space, let x_{0} be some chosen point of X, and let π_{1}(X, x_{0}) be the set of all based homotopy classes of loops based at the point x_{0}. Then π_{1}(X, x_{0}) is a group, the group multiplication on π_{1}(X, x_{0}) being defined according to the rule [γ_{1}][γ_{2}] = [γ_{1}.γ_{2}] for all loops γ_{1} and γ_{2} based at x_{0}.

Proof First we show that the group operation on π_{1}(X, x_{0}) is well-defined. Let γ_{1}, γ′_{1}, γ_{2} and γ′_{2} be loops in X based at the point x_{0}. Suppose that [γ_{1}] = [γ′_{1}] and [γ_{2}] = [γ′_{2}]. Let the map F: [0, 1] × [0, 1] → X be defined by

F (t, τ) = \{ \begin{matrix}F_{1}(2t, τ) & if 0 ≤ t ≤ \frac{1}{2}, \\ F_{2}(2t − 1, τ) & if \frac{1}{2} ≤ t ≤ 1, \end{matrix}

where F_{1}: [0, 1] × [0, 1] → X is a homotopy between γ_{1} and γ′_{1}, F_{2}: [0, 1] × [0, 1] → X is a homotopy between γ_{2} and γ′_{2}, and where the homotopies F_{1} and F_{2} map (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]. Then F is itself a homotopy from γ_{1}.γ_{2} to γ′_{1}.γ′_{2}, and maps (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]. Thus [γ_{1}.γ_{2}] = [γ′_{1}.γ′_{2}], showing that the group operation on π_{1}(X, x_{0}) is well-defined.

Next we show that the group operation on π_{1}(X, x_{0}) is associative. Let γ_{1}, γ_{2} and γ_{3} be loops based at x_{0}, and let α = (γ_{1}.γ_{2}).γ_{3}. Then γ_{1}.(γ_{2}.γ_{3}) = α∘θ, where

θ(t) = \{ \begin{matrix}\frac{1}{2}t & if 0 ≤ t ≤ \frac{1}{2}; \\ t − \frac{1}{4} & if \frac{1}{2} ≤ t ≤ \frac{3}{4}; \\ 2t − 1 & if \frac{3}{4} ≤ t ≤ 1. \end{matrix}

Thus the map G: [0, 1]×[0, 1] → X defined by G(t, τ) = α((1−τ)t+τθ(t)) is a homotopy between (γ_{1}.γ_{2}).γ_{3} and γ_{1}.(γ_{2}.γ_{3}), and moreover this homotopy maps (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]. It follows that (γ_{1}.γ_{2}).γ_{3} ≃ γ_{1}.(γ_{2}.γ_{3}) rel {0, 1} and hence ([γ_{1}][γ_{2}])[γ_{3}] = [γ_{1}]([γ_{2}][γ_{3}]). This shows that the group operation on π_{1}(X, x_{0}) is associative.

Let ε: [0, 1] → X denote the constant loop at x_{0}, defined by ε(t) = x_{0} for all t ∈ [0, 1]. Then ε.γ = γ ∘ θ_{0} and γ.ε = γ ∘ θ_{1} for any loop γ based at x_{0}, where

θ_{0}(t) = \{ \begin{matrix}0 & if 0 ≤ t ≤ \frac{1}{2}, \\ 2t − 1 & if \frac{1}{2} ≤ t ≤ 1, \end{matrix} θ_{1}(t) = \{ \begin{matrix}2t & if 0 ≤ t ≤ \frac{1}{2}, \\ 1 & if \frac{1}{2} ≤ t ≤ 1, \end{matrix}

for all t ∈ [0, 1]. But the continuous map (t, τ) ↦ γ((1 − τ)t + τθ_{j}(t)) is a homotopy between γ and γ ∘ θ_{j} for j = 0, 1 which sends (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]. Therefore ε.γ ≃ γ ≃ γ.ε rel {0, 1}, and hence [ε][γ] = [γ] = [γ][ε]. We conclude that [ε] represents the identity element of π_{1}(X, x_{0}).

It only remains to verify the existence of inverses. Now the map K: [0, 1]× [0, 1] → X defined by

K(t, τ) = \{ \begin{matrix}γ(2τt) & if 0 ≤ t ≤ \frac{1}{2}; \\ γ(2τ(1 − t)) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

is a homotopy between the loops γ.γ^{−1} and ε, and moreover this homotopy sends (0, τ) and (1, τ) to x_{0} for all τ ∈ [0, 1]. Therefore γ.γ^{−1} ≃ ε rel{0, 1}, and thus [γ][γ^{−1}] = [γ.γ^{−1}] = [ε]. On replacing γ by γ^{−1}, we see also that [γ^{−1}][γ] = [ε], and thus [γ^{−1}] = [γ]^{−1}, as required.

Let x_{0} be a point of some topological space X. The group π_{1}(X, x_{0}) is referred to as the fundamental group of X based at the point x_{0}.

Let f: X → Y be a continuous map between topological spaces X and Y, and let x_{0} be a point of X. Then f induces a homomorphism f_{#}: π_{1}(X, x_{0}) → π_{1}(Y, f(x_{0})), where f_{#}([γ]) = [f ∘ γ] for all loops γ: [0, 1] → X based at x_{0}. If x_{0}, y_{0} and z_{0} are points belonging to topological spaces X, Y and Z, and if f: X → Y and g: Y → Z are continuous maps satisfying f(x_{0}) = y_{0} and g(y_{0}) = z_{0}, then the induced homomorphisms f_{#}: π_{1}(X, x_{0}) → π_{1}(Y, y_{0}) and g_{#}: π_{1}(Y, x_{0}) → π_{1}(Z, z_{0}) satisfy g_{#} ∘ f_{#} = (g ∘ f)_{#}. It follows easily from this that any homeomorphism of topological spaces induces a corresponding isomorphism of fundamental groups, and thus the fundamental group is a topological invariant.

3.2 Simply-Connected Topological Spaces

Definition A topological space X is said to be simply-connected if it is pathconnected, and any continuous map f: ∂D → X mapping the boundary circle ∂D of a closed disc D into X can be extended continuously over the whole of the disk.

Example ℝ^{n} is simply-connected for all n. Indeed any continuous map f: ∂D → ℝ^{n} defined over the boundary ∂D of the closed unit disk D can be extended to a continuous map F: D → ℝ^{n} over the whole disk by setting F (rx) = rf(x) for all x ∈ ∂D and r ∈ [0, 1].

Let E be a topological space that is homeomorphic to the closed disk D, and let ∂E = h(∂D), where ∂D is the boundary circle of the disk D and h: D → E is a homeomorphism from D to E. Then any continuous map g: ∂E → X mapping ∂E into a simply-connected space X extends continuously to the whole of E. Indeed there exists a continuous map F: D → X which extends g ∘ h: ∂D → X, and the map F ∘ h^{−1}: E → X then extends the map g.

Theorem 3.2 A path-connected topological space X is simply-connected if and only if π_{1}(X, x) is trivial for all x ∈ X.

Proof Suppose that the space X is simply-connected. Let γ: [0, 1] → X be a loop based at some point x of X. Now the unit square is homeomorphic to the unit disk, and therefore any continuous map defined over the boundary of the square can be continuously extended over the whole of the square. It follows that there exists a continuous map F: [0, 1] × [0, 1] → X such that F (t, 0) = γ(t) and F (t, 1) = x for all t ∈ [0, 1], and F (0, τ) = F (1, τ) = x for all τ ∈ [0, 1]. Thus γ ≃ ε_{x} rel{0, 1}, where ε_{x} is the constant loop at x, and hence [γ] = [ε_{x}] in π_{1}(X, x). This shows that π_{1}(X, x) is trivial.

Conversely suppose that X is path-connected and π_{1}(X, x) is trivial for all x ∈ X. Let f: ∂D → X be a continuous function defined on the boundary circle ∂D of the closed unit disk D in ℝ^{2}. We must show that f can be extended continuously over the whole of D. Let x = f(1, 0). There exists a continuous map G: [0, 1]×[0, 1] → X such that G(t, 0) = f(cos(2πt), sin(2πt)) and G(t, 1) = x for all t ∈ [0, 1] and G(0, τ) = G(1, τ) = x for all τ ∈ [0, 1], since π_{1}(X, x) is trivial. Moreover G(t_{1}, τ_{1}) = G(t_{2}, τ_{2}) whenever q(t_{1}, τ_{1}) = q(t_{2}, τ_{2}), where

q(t, τ) = ((1 − τ) cos(2πt) + τ, (1 − τ) sin(2πt))

for all t, τ ∈ [0, 1]. It follows that there is a well-defined function F: D → X such that F ∘ q = G. However q: [0, 1] × [0, 1] → D is a continuous surjection from a compact space to a Hausdorff space and is therefore an identification map. It follows that F: D → X is continuous (since a basic property of identification maps ensures that a function F: D → X is continuous if and only if F ∘ q: [0, 1] × [0, 1] → X is continuous). Moreover F: D → X extends the map f. We conclude that the space X is simply-connected, as required.

One can show that, if two points x_{1} and x_{2} in a topological space X can be joined by a path in X then π_{1}(X, x_{1}) and π_{1}(X, x_{2}) are isomorphic. On combining this result with Theorem 3.2, we see that a path-connected topological space X is simply-connected if and only if π_{1}(X, x) is trivial for some x ∈ X.

Theorem 3.3 Let X be a topological space, and let U and V be open subsets of X, with U ∪V = X. Suppose that U and V are simply-connected, and that U ∩ V is non-empty and path-connected. Then X is itself simply-connected.

Proof We must show that any continuous function f: ∂D → X defined on the unit circle ∂D can be extended continuously over the closed unit disk D. Now the preimages f^{−1}(U) and f^{−1}(V) of U and V are open in ∂D (since f is continuous), and ∂D = f^{−1}(U) ∪ f^{−1}(V). It follows from the Lebesgue Lemma that there exists some δ > 0 such that any arc in ∂D whose length is less than δ is entirely contained in one or other of the sets f^{−1}(U) and f^{−1}(V). Choose points z_{1}, z_{2}, …, z_{n} around ∂D such that the distance from z_{i} to z_{i+1} is less than δ for i = 1, 2, …, n−1 and the distance from z_{n} to z_{1} is also less than δ. Then, for each i, the short arc joining z_{i−1} to z_{i} is mapped by f into one or other of the open sets U and V.

Let x_{0} be some point of U ∩V. Now the sets U, V and U ∩V are all pathconnected. Therefore we can choose paths α_{i}: [0, 1] → X for i = 1, 2, …, n such that α_{i}(0) = x_{0}, α_{i}(1) = f(z_{i}), α_{i}([0, 1]) ⊂ U whenever f(z_{i}) ∈ U, and α_{i}([0, 1]) ⊂ V whenever f(z_{i}) ∈ V. For convenience let α_{0} = α_{n}.

Now, for each i, consider the sector T_{i} of the closed unit disk bounded by the line segments joining the centre of the disk to the points z_{i−1} and z_{i} and by the short arc joining z_{i−1} to z_{i}. Now this sector is homeomorphic to the closed unit disk, and therefore any continuous function mapping the boundary ∂T_{i} of T_{i} into a simply-connected space can be extended continuously over the whole of T_{i}. In particular, let F_{i} be the function on ∂T_{i} defined by

F_{i}(z) = \{ \begin{matrix}f(z) & if z ∈ T_{i} ∩ ∂D, \\ α_{i−1}(t) & if z = tz_{i−1} for any t ∈ [0, 1], \\ α_{i}(t) & if z = tz_{i} for any t ∈ [0, 1], \end{matrix}

Note that F_{i}(∂T_{i}) ⊂ U whenever the short arc joining z_{i−1} to z_{i} is mapped by f into U, and F_{i}(∂T_{i}) ⊂ V whenever this short arc is mapped into V. But U and V are both simply-connected. It follows that each of the functions F_{i} can be extended continuously over the whole of the sector T_{i}. Moreover the functions defined in this fashion on each of the sectors T_{i} agree with one another wherever the sectors intersect, and can therefore be pieced together to yield a continuous map defined over the the whole of the closed disk D which extends the map f, as required.

Example The n-dimensional sphere S^{n} is simply-connected for all n > 1, where S^{n} = \{x ∈ ℝ^{n+1} : |x| = 1\}. Indeed let U = \{x ∈ S^{n} : x_{n+1} > −\frac{1}{2}\} and V = \{x ∈ S^{n} : x_{n+1} < \frac{1}{2}\}. Then U and V are homeomorphic to an n-dimensional ball, and are therefore simply-connected. Moreover U ∩ V is path-connected, provided that n > 1. It follows that S^{n} is simply-connected for all n > 1.

3.3 The Fundamental Group of the Circle

Theorem 3.4 π_{1}(S^{1}, b) ≅ ℤ for any b ∈ S^{1}.

Proof We regard S^{1} as the unit circle in ℝ^{2}. Without loss of generality, we can take b = (1, 0). Now the map p: ℝ → S^{1} which sends t ∈ ℝ to (cos 2πt, sin 2πt) is a covering map, and b = p(0). Moreover p(t_{1}) = p(t_{2}) if and only if t_{1} − t_{2} is an integer; in particular p(t) = b if and only if t is an integer.

Let α and β be loops in S^{1} based at b, and let α̃ and β̃ be paths in ℝ that satisfy p ∘ α̃ = α and p ∘ β̃ = β. Suppose that α and β represent the same element of π_{1}(S^{1}, b). Then there exists a homotopy F: [0, 1] × [0, 1] → S^{1} such that F (t, 0) = α(t) and F (t, 1) = β(t) for all t ∈ [0, 1], and F (0, τ) = F (1, τ) = b for all τ ∈ [0, 1]. It follows from the Monodromy Theorem (Theorem 2.6) that this homotopy lifts to a continuous map G: [0, 1]×[0, 1] → ℝ satisfying p ∘ G = F. Moreover G(0, τ) and G(1, τ) are integers for all τ ∈ [0, 1], since p(G(0, τ)) = b = p(G(1, τ)). Also G(t, 0)−α̃(t) and G(t, 1)− β̃(t) are integers for all t ∈ [0, 1], since p(G(t, 0)) = α(t) = p(α̃(t)) and p(G(t, 1)) = β(t) = p(β̃(t)). Now any continuous integer-valued function on [0, 1] is constant, by the Intermediate Value Theorem. In particular the functions sending τ ∈ [0, 1] to G(0, τ) and G(1, τ) are constant, as are the functions sending t ∈ [0, 1] to G(t, 0) − α̃(t) and G(t, 1) − β̃(t). Thus

G(0, 0) = G(0, 1), G(1, 0) = G(1, 1),

G(1, 0) − α̃(1) = G(0, 0) − α̃(0), G(1, 1) − β̃(1) = G(0, 1) − β̃(0).

On combining these results, we see that

α̃(1) − α̃(0) = G(1, 0) − G(0, 0) = G(1, 1) − G(0, 1) = β̃(1) − β̃(0).

We conclude from this that there exists a well-defined function λ: π_{1}(S^{1}, b) → ℤ characterized by the property that λ([α]) = α̃(1)−α̃(0) for all loops α based at b, where α̃: [0, 1] → ℝ is any path in ℝ satisfying p ∘ α̃ = α.

Next we show that λ is a homomorphism. Let α and β be any loops based at b, and let α̃ and β̃ be lifts of α and β. The element [α][β] of π_{1}(S^{1}, b) is represented by the product path α.β, where

(α.β)(t) = \{ \begin{matrix}α(2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ β(2t − 1) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

Define a continuous path σ: [0, 1] → ℝ by

σ(t) = \{ \begin{matrix}α̃(2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ β̃(2t − 1) + α̃(1) − β̃(0) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

(Note that σ(t) is well-defined when t = \frac{1}{2}.) Then p ∘ σ = α.β and thus

λ([α][β]) = λ([α.β]) = σ(1) − σ(0) = α̃(1) − α̃(0) + β̃(1) − β̃(0)
= λ([α]) + λ([β]).

Thus λ: π_{1}(S^{1}, b) → ℤ is a homomorphism.

Now suppose that λ([α]) = λ([β]). Let F: [0, 1] × [0, 1] → S^{1} be the homotopy between α and β defined by

F (t, τ) = p ((1 − τ)α̃(t) + τβ̃(t)) ,

where α̃ and β̃ are the lifts of α and β respectively starting at 0. Now β̃(1) = λ([β]) = λ([α]) = α̃(1), and β̃(0) = α̃(0) = 0. Therefore F (0, τ) = b = p(α̃(1)) = F (1, τ) for all τ ∈ [0, 1]. Thus α ≃ β rel {0, 1}, and therefore [α] = [β]. This shows that λ: π_{1}(S^{1}, b) → ℤ is injective.

The homomorphism λ is surjective, since n = λ([γ_{n}]) for all n ∈ ℤ, where the loop γ_{n}: [0, 1] → S^{1} is given by γ_{n}(t) = p(nt) = (cos 2πnt, sin 2πnt) for all t ∈ [0, 1]. We conclude that λ: π_{1}(S^{1}, b) → ℤ is an isomorphism.

We now show that every continuous map from the closed disk D to itself has at least one fixed point. This is the two-dimensional version of the Brouwer Fixed Point Theorem.

Theorem 3.5 Let f: D → D be a continuous map which maps the closed disk D into itself. Then f(x_{0}) = x_{0} for some x_{0} ∈ D.

Proof Let ∂D denote the boundary circle of D. The inclusion map i: ∂D ↪ D induces a corresponding homomorphism i_{#}: π_{1}(∂D, b) → π_{1}(D, b) of fundamental groups for any b ∈ ∂D.

Suppose that it were the case that the map f has no fixed point in D. Then one could define a continuous map r: D → ∂D as follows: for each x ∈ D, let r(x) be the point on the boundary ∂D of D obtained by continuing the line segment joining f(x) to x beyond x until it intersects ∂D at the point r(x). Note that r|∂D is the identity map of ∂D.

Let r_{#}: π_{1}(D, b) → π_{1}(∂D, b) be the homomorphism of fundamental groups induced by r: D → ∂D. Now (r ∘ i)_{#}: π_{1}(∂D, b) → π_{1}(∂D, b) is the identity isomorphism of π_{1}(∂D, b), since r ∘ i: ∂D → ∂D is the identity map. But it follows directly from the definition of induced homomorphisms that (r ∘ i)_{#} = r_{#} ∘ i_{#}. Therefore i_{#}: π_{1}(∂D, b) → π_{1}(D, b) is injective, and r_{#}: π_{1}(D, b) → π_{1}(∂D, b) is surjective. But this is impossible, since π_{1}(∂D, b) ≅ ℤ (Theorem 3.4) and π_{1}(D, b) is the trivial group. This contradiction shows that the continuous map f: D → D must have at least one fixed point.

4 Simplicial Complexes

4.1 Geometrical Independence

Definition Points v_{0}, v_{1}, …, v_{q} in some Euclidean space ℝ^{k} are said to be geometrically independent (or affine independent) if the only solution of the linear system

\{ \begin{matrix}∑_{j=0}^{q} λ_{j}v_{j} & = & 0, \\ ∑_{j=0}^{q} λ_{j} & = & 0 \end{matrix}

is the trivial solution λ_{0} = λ_{1} = ⋯ = λ_{q} = 0.

It is straightforward to verify that v_{0}, v_{1}, …, v_{q} are geometrically independent if and only if the vectors v_{1} − v_{0}, v_{2} − v_{0}, …, v_{q} − v_{0} are linearly independent. It follows from this that any set of geometrically independent points in ℝ^{k} has at most k + 1 elements. Note also that if a set consists of geometrically independent points in ℝ^{k}, then so does every subset of that set.

Definition A q-simplex in ℝ^{k} is defined to be a set of the form

{∑_{j=0}^{q} t_{j}v_{j} : 0 ≤ t_{j} ≤ 1 for j = 0, 1, …, q and ∑_{j=0}^{q} t_{j} = 1} ,

where v_{0}, v_{1}, …, v_{q} are geometrically independent points of ℝ^{k}. The points v_{0}, v_{1}, …, v_{q} are referred to as the vertices of the simplex. The non-negative integer q is referred to as the dimension of the simplex.

Note that a 0-simplex in ℝ^{k} is a single point of ℝ^{k}, a 1-simplex in ℝ^{k} is a line segment in ℝ^{k}, a 2-simplex is a triangle, and a 3-simplex is a tetrahedron.

Let σ be a q-simplex in ℝ^{k} with vertices v_{0}, v_{1}, …, v_{q}. If x is a point of σ then there exist real numbers t_{0}, t_{1}, …, t_{q} such that

∑_{j=0}^{q} t_{j}v_{j} = x, ∑_{j=0}^{q} t_{j} = 1 and 0 ≤ t_{j} ≤ 1 for j = 0, 1, …, q.

Moreover t_{0}, t_{1}, …, t_{q} are uniquely determined: if ∑_{j=0}^{q} s_{j}v_{j} = ∑_{j=0}^{q} t_{j}v_{j} and

∑_{j=0}^{q} s_{j} = 1 = ∑_{j=0}^{q} t_{j}, then ∑_{j=0}^{q}(t_{j} − s_{j})v_{j} = 0 and ∑_{j=0}^{q}(t_{j} − s_{j}) = 0, hence

t_{j} − s_{j} = 0 for all j, since v_{0}, v_{1}, …, v_{q} are geometrically independent. We refer to t_{0}, t_{1}, …, t_{q} as the barycentric coordinates of the point x of σ.

Lemma 4.1 Let q be a non-negative integer, let σ be a q-simplex in ℝ^{m}, and let τ be a q-simplex in ℝ^{n}, where m ≥ q and n ≥ q. Then σ and τ are homeomorphic.

Proof Let v_{0}, v_{1}, …, v_{q} be the vertices of σ, and let w_{0}, w_{1}, …, w_{q} be the vertices of τ. The required homeomorphism h: σ → τ is given by

h (∑_{j=0}^{q} t_{j}v_{j}) = ∑_{j=0}^{q} t_{j}w_{j}

for all t_{0}, t_{1}, …, t_{q} satisfying 0 ≤ t_{j} ≤ 1 for j = 0, 1, …, q and ∑_{j=0}^{q} t_{j} = 1.

A homeomorphism between two q-simplices defined as in the above proof is referred to as a simplicial homeomorphism.

4.2 Simplicial Complexes in Euclidean Spaces

Definition Let σ and τ be simplices in ℝ^{k}. We say that τ is a face of σ if the set of vertices of τ is a subset of the set of vertices of σ. A face of σ is said to be a proper face if it is not equal to σ itself. An r-dimensional face of σ is referred to as an r-face of σ. A 1-dimensional face of σ is referred to as an edge of σ.

Note that any simplex is a face of itself. Also the vertices and edges of any simplex are by definition faces of the simplex.

Definition A finite collection K of simplices in ℝ^{k} is said to be a simplicial complex if the following two conditions are satisfied:—

• if σ is a simplex belonging to K then every face of σ also belongs to K,

• if σ_{1} and σ_{2} are simplices belonging to K then either σ_{1} ∩ σ_{2} = ∅ or else σ_{1} ∩ σ_{2} is a common face of both σ_{1} and σ_{2}.

The dimension of a simplicial complex K is the greatest non-negative integer n with the property that K contains an n-simplex. The union of all the simplices of K is a compact subset |K| of ℝ^{k} referred to as the polyhedron of K. (The polyhedron is compact since it is both closed and bounded in ℝ^{k}.)

Example Let K_{σ} consist of some n-simplex σ together with all of its faces. Then K_{σ} is a simplicial complex of dimension n, and |K_{σ}| = σ.

Lemma 4.2 Let K be a simplicial complex, and let X be a topological space. A function f: |K| → X is continuous on the polyhedron |K| of K if and only if the restriction of f to each simplex of K is continuous on that simplex.

Proof If a topological space can be expressed as a finite union of closed subsets, then a function is continuous on the whole space if and only if its restriction to each of the closed subsets is continuous on that closed set. The required result is a direct application of this general principle.

We shall denote by Vert K the set of vertices of a simplicial complex K (i.e., the set consisting of all vertices of all simplices belonging to K). A collection of vertices of K is said to span a simplex of K if these vertices are the vertices of some simplex belonging to K.

Definition Let K be a simplicial complex in ℝ^{k}. A subcomplex of K is a collection L of simplices belonging to K with the following property:—

• if σ is a simplex belonging to L then every face of σ also belongs to L.

Note that every subcomplex of a simplicial complex K is itself a simplicial complex.

Definition Let v_{0}, v_{1}, …, v_{q} be the vertices of a q-simplex σ in some Euclidean space ℝ^{k}. We define the interior of the simplex σ to be the set of all

points of σ that are of the form ∑_{j=0}^{q} t_{j}v_{j}, where t_{j} > 0 for j = 0, 1, …, q and

∑_{j=0}^{q} t_{j} = 1. One can readily verify that the interior of the simplex σ consists

of all points of σ that do not belong to any proper face of σ. (Note that, if σ ∈ ℝ^{k}, then the interior of a simplex defined in this fashion will not coincide with the topological interior of σ unless dim σ = k.)

Note that any point of a simplex σ belongs to the interior of a unique face of σ. Indeed let v_{0}, v_{1}, …, v_{q} be the vertices of σ, and let x ∈ σ. Then

x = ∑_{j=0}^{q} t_{j}v_{j}, where 0 ≤ t_{j} ≤ 1 for j = 0, 1, …, q and ∑_{j=0}^{q} t_{j} = 1. The

unique face of σ containing x in its interior is then the face spanned by those vertices v_{j} for which t_{j} > 0.

Lemma 4.3 Let K be a finite collection of simplices in some Euclidean space ℝ^{k}, and let |K| be the union of all the simplices in K. Then K is a simplicial complex (with polyhedron |K|) if and only if the following two conditions are satisfied:—

• K contains the faces of its simplices,

• every point of |K| belongs to the interior of a unique simplex of K.

Proof Suppose that K is a simplicial complex. Then K contains the faces of its simplices. We must show that every point of |K| belongs to the interior of a unique simplex of K. Let x ∈ |K|. Then x belongs to the interior of a face σ of some simplex of K (since every point of a simplex belongs to the interior of some face). But then σ ∈ K, since K contains the faces of all its simplices. Thus x belongs to the interior of at least one simplex of K.

Suppose that x were to belong to the interior of two distinct simplices σ and τ of K. Then x would belong to some common face σ ∩ τ of σ and τ (since K is a simplicial complex). But this common face would be a proper face of one or other of the simplices σ and τ (since σ ≠ τ), contradicting the fact that x belongs to the interior of both σ and τ. We conclude that the simplex σ of K containing x in its interior is uniquely determined, as required.

Conversely, we must show that any collection of simplices satisfying the given conditions is a simplicial complex. Since K contains the faces of all its simplices, it only remains to verify that if σ and τ are any two simplices of K with non-empty intersection then σ ∩ τ is a common face of σ and τ.

Let x ∈ σ ∩ τ. Then x belongs to the interior of a unique simplex ω of K. However any point of σ or τ belongs to the interior of a unique face of that simplex, and all faces of σ and τ belong to K. It follows that ω is a common face of σ and τ, and thus the vertices of ω are vertices of both σ and τ. We deduce that the simplices σ and τ have vertices in common, and that every point of σ ∩ τ belongs to the common face ρ of σ and τ spanned by these common vertices. But this implies that σ ∩ τ = ρ, and thus σ ∩ τ is a common face of both σ and τ, as required.

Definition A triangulation (K, h) of a topological space X consists of a simplicial complex K in some Euclidean space, together with a homeomorphism h: |K| → X mapping the polyhedron |K| of K onto X.

The polyhedron of a simplicial complex is a compact Hausdorff space. Thus if a topological space admits a triangulation then it must itself be a compact Hausdorff space.

Lemma 4.4 Let X be a Hausdorff topological space, let K be a simplicial complex, and let h: |K| → X be a bijection mapping |K| onto X. Suppose that the restriction of h to each simplex of K is continuous on that simplex. Then the map h: |K| → X is a homeomorphism, and thus (K, h) is a triangulation of X.

Proof Each simplex of K is a closed subset of |K|, and the number of simplices of K is finite. It follows from Lemma 4.2 that h: |K| → X is continuous. Also the polyhedron |K| of K is a compact topological space. But every continuous bijection from a compact topological space to a Hausdorff space is a homeomorphism. Thus (K, h) is a triangulation of X.

4.3 Simplicial Maps

Definition A simplicial map φ: K → L between simplicial complexes K and L is a function φ: Vert K → Vert L from the vertex set of K to that of L such that φ(v_{0}), φ(v_{1}), …, φ(v_{q}) span a simplex belonging to L whenever v_{0}, v_{1}, …, v_{q} span a simplex of K.

Note that a simplicial map φ: K → L between simplicial complexes K and L can be regarded as a function from K to L: this function sends a simplex σ of K with vertices v_{0}, v_{1}, …, v_{q} to the simplex φ(σ) of L spanned by the vertices φ(v_{0}), φ(v_{1}), …, φ(v_{q}).

A simplicial map φ: K → L also induces in a natural fashion a continuous map φ: |K| → |L| between the polyhedra of K and L, where

φ (∑_{j=0}^{q} t_{j}v_{j}) = ∑_{j=0}^{q} t_{j}φ(v_{j})

whenever 0 ≤ t_{j} ≤ 1 for j = 0, 1, …, q, ∑_{j=0}^{q} t_{j} = 1, and v_{0}, v_{1}, …, v_{q} span a

simplex of K. The continuity of this map follows immediately from a straightforward application of Lemma 4.2. Note that the interior of a simplex σ of K is mapped into the interior of the simplex φ(σ) of L.

There are thus three equivalent ways of describing a simplicial map: as a function between the vertex sets of two simplicial complexes, as a function from one simplicial complex to another, and as a continuous map between the polyhedra of two simplicial complexes. In what follows, we shall describe a simplicial map using the representation that is most appropriate in the given context.

4.4 Barycentric Subdivision of a Simplicial Complex

Let σ be a q-simplex in ℝ^{k} with vertices v_{0}, v_{1}, …, v_{q}. The barycentre of σ is defined to be the point

σ̂ = \frac{1}{q + 1}(v_{0} + v_{1} + ⋯ + v_{q}).

Let σ and τ be simplices in some Euclidean space. If σ is a proper face of τ then we denote this fact by writing σ < τ.

A simplicial complex K_{1} is said to be a subdivision of a simplicial complex K if |K_{1}| = |K| and each simplex of K_{1} is contained in a simplex of K.

Definition Let K be a simplicial complex in some Euclidean space ℝ^{k}. The first barycentric subdivision K′ of K is defined to be the collection of simplices in ℝ^{k} whose vertices are σ̂_{0}, σ̂_{1}, …, σ̂_{r} for some sequence σ_{0}, σ_{1}, …, σ_{r} of simplices of K with σ_{0} < σ_{1} < ⋯ < σ_{r}. Thus the set of vertices of K′ is the set of all the barycentres of all the simplices of K.

Note that every simplex of K′ is contained in a simplex of K. Indeed if σ_{0}, σ_{1}, …, σ_{r} ∈ K satisfy σ_{0} < σ_{1} < ⋯ < σ_{r} then the simplex of K′ spanned by σ̂_{0}, σ̂_{1}, …, σ̂_{r}, is contained in the simplex σ_{r} of K.

Proposition 4.5 Let K be a simplicial complex in some Euclidean space, and let K′ be the first barycentric subdivision of K. Then K′ is itself a simplicial complex, and |K′| = |K|.

Proof We prove the result by induction on the number of simplices in K. The result is clear when K consists of a single simplex, since that simplex must then be a point and therefore K′ = K. We prove the result for a simplicial complex K, assuming that it holds for all complexes with fewer simplices.

It is clear from the definition of the barycentric subdivision K′ that any face of a simplex of K′ must itself belong to K′. We must verify that any two simplices of K′ are disjoint or else intersect in a common face.

Choose a simplex σ of K for which dim σ = dim K, and let L = K \ {σ}. Then L is a subcomplex of K, since σ is not a proper face of any simplex of K. Now L has fewer simplices than K. It follows from the induction hypothesis that L′ is a simplicial complex and |L′| = |L|. Also it follows from the definition of K′ that K′ consists of the following simplices:—

• the simplices of L′,

• the barycentre σ̂ of σ,

• simplices σ̂ρ whose vertex set is obtained by adjoining σ̂ to the vertex set of some simplex ρ of L′, where the vertices of ρ are barycentres of proper faces of σ.

By checking all possible intersections of simplices of the above types, it is easy to verify that any two simplices of K′ intersect in a common face. Indeed any two simplices of L′ intersect in a common face, since L′ is a simplicial complex. If ρ_{1} and ρ_{2} are simplices of L′ whose vertices are barycentres of proper faces of σ, then ρ_{1} ∩ ρ_{2} is a common face of ρ_{1} and ρ_{2} which is of this type, and σ̂ρ_{1} ∩ σ̂ρ_{2} = σ̂(ρ_{1} ∩ ρ_{2}). Thus σ̂ρ_{1} ∩ σ̂ρ_{2} is a common face of σ̂ρ_{1} and σ̂ρ_{2}. Also any simplex τ of L′ is disjoint from the barycentre σ̂ of σ, and σ̂ρ ∩ τ = ρ ∩ τ. We conclude that K′ is indeed a simplicial complex.

It remains to verify that |K′| = |K|. Now |K′| ⊂ |K|, since every simplex of K′ is contained in a simplex of K. Let x be a point of the chosen simplex σ. Then there exists a point y belonging to a proper face of σ and some t ∈ [0, 1] such that x = (1−t)σ̂+ty. But then y ∈ |L|, and |L| = |L′| by the induction hypothesis. It follows that y ∈ ρ for some simplex ρ of L′ whose vertices are barycentres of proper faces of σ. But then x ∈ σ̂ρ, and therefore x ∈ |K′|. Thus |K| ⊂ |K′|, and hence |K′| = |K|, as required.

We define (by induction on j) the jth barycentric subdivision K^{(j)} of K to be the first barycentric subdivision of K^{(j−1)} for each j > 1.

Lemma 4.6 Let σ be a q-simplex and let τ be a face of σ. Let σ̂ and τ̂ be the barycentres of σ and τ respectively. If all the 1-simplices (edges) of σ have length not exceeding d for some d > 0 then

|σ̂ − τ̂| ≤ \frac{qd}{q + 1}.

Proof Let v_{0}, v_{1}, …, v_{q} be the vertices of σ. Let x and y be points of σ.

We can write y = ∑_{j=0}^{q} t_{j}v_{j}, where 0 ≤ t_{i} ≤ 1 for i = 0, 1, …, q and ∑_{j=0}^{q} t_{j} = 1.

Now

|x − y| = |∑_{i=0}^{q} t_{i}(x − v_{i})| ≤ ∑_{i=0}^{q} t_{i}|x − v_{i}|
≤ maximum (|x − v_{0}|, |x − v_{1}|, …, |x − v_{q}|) .

Applying this result with x = σ̂ and y = τ̂, we find that

|σ̂ − τ̂| ≤ maximum (|σ̂ − v_{0}|, |σ̂ − v_{1}|, …, |σ̂ − v_{q}|) .

But

σ̂ = \frac{1}{q + 1}v_{i} + \frac{q}{q + 1}z_{i}

for i = 0, 1, …, q, where z_{i} is the barycentre of the (q − 1)-face of σ opposite to v_{i}, given by

z_{i} = \frac{1}{q} ∑_{j≠i} v_{j}.

Moreover z_{i} ∈ σ. It follows that

|σ̂ − v_{i}| = \frac{q}{q + 1}|z_{i} − v_{i}| ≤ \frac{qd}{q + 1}

for i = 1, 2, …, q, and thus

|σ̂ − τ̂| ≤ maximum (|σ̂ − v_{0}|, |σ̂ − v_{1}|, …, |σ̂ − v_{q}|) ≤ \frac{qd}{q + 1},

as required.

The mesh μ(K) of a simplicial complex K is the length of the longest edge of K.

Lemma 4.7 Let K be a simplicial complex in ℝ^{k} for some k, and let n be the dimension of K. Let K′ be the first barycentric subdivision of K. Then

μ(K′) ≤ \frac{n}{n + 1}μ(K).

Proof A 1-simplex of K′ is of the form (τ̂, σ̂), where σ is a q-simplex of K for some q ≤ n and τ is a proper face of σ. Then

|τ̂ − σ̂| ≤ \frac{q}{q + 1}μ(K) ≤ \frac{n}{n + 1}μ(K)

by Lemma 4.6, as required.

It follows directly from the above lemma that lim_{j→+∞} μ(K^{(j)}) = 0, where

K^{(j)} is the jth barycentric subdivision of K.

4.5 The Simplicial Approximation Theorem

Definition Let f: |K| → |L| be a continuous map between the polyhedra of simplicial complexes K and L. A simplicial map s: K → L is said to be a simplicial approximation to f if, for each x ∈ |K|, s(x) is an element of the unique simplex of L which contains f(x) in its interior.

Note that if s: K → L is a simplicial approximation to f: |K| → |L| then s and f are homotopic. Indeed the map from |K| × [0, 1] to |L| sending (x, t) to (1 − t)f(x) + ts(x) is a well-defined homotopy between f and s.

Definition Let K be a simplicial complex, and let x ∈ |K|. The star st_{K}(x) of x in K is the union of the interiors of all simplices of K that contain the point x.

Lemma 4.8 Let K be a simplicial complex and let x ∈ |K|. Then the star st_{K}(x) of x is open in |K|, and x ∈ st_{K}(x).

Proof Every point of |K| belongs to the interior of a unique simplex of K (Lemma 4.3). It follows that the complement |K| \ st_{K}(x) of st_{K}(x) in |K| is the union of the interiors of those simplices of K that do not contain the point x. But if a simplex of K does not contain the point x, then the same is true of its faces. Moreover the union of the interiors of all the faces of some simplex is the simplex itself. It follows that |K| \ st_{K}(x) is the union of all simplices of K that do not contain the point x. But each simplex of K is closed in |K|. It follows that |K| \ st_{K}(x) is a finite union of closed sets, and is thus itself closed in |K|. We deduce that st_{K}(x) is open in |K|. Also x ∈ st_{K}(x), since x belongs to the interior of at least one simplex of K.

Proposition 4.9 A function s: Vert K → Vert L between the vertex sets of simplicial complexes K and L is a simplicial map, and a simplicial approximation to some continuous map f: |K| → |L|, if and only if f (st_{K}(v)) ⊂ st_{L} (s(v)) for all vertices v of K.

Proof Let s: K → L be a simplicial approximation to f: |K| → |L|, let v be a vertex of K, and let x ∈ st_{K}(v). Then x and f(x) belong to the interiors of unique simplices σ ∈ K and τ ∈ L. Moreover v must be a vertex of σ, by definition of st_{K}(v). Now s(x) must belong to τ (since s is a simplicial approximation to the map f), and therefore s(x) must belong to the interior of some face of τ. But s(x) must belong to the interior of s(σ), since x is in the interior of σ. It follows that s(σ) must be a face of τ, and therefore s(v) must be a vertex of τ. Thus f(x) ∈ st_{L}(s(v)). We conclude that if s: K → L is a simplicial approximation to f: |K| → |L|, then f (st_{K}(v)) ⊂ st_{L} (s(v)).

Conversely let s: Vert K → Vert L be a function with the property that f (st_{K}(v)) ⊂ st_{L} (s(v)) for all vertices v of K. Let x be a point in the interior of some simplex of K with vertices v_{0}, v_{1}, …, v_{q}. Then x ∈ st_{K}(v_{j}) and hence f(x) ∈ st_{L} (s(v_{j})) for j = 0, 1, …, q. It follows that each vertex s(v_{j}) must be a vertex of the unique simplex τ ∈ L that contains f(x) in its interior. In particular, s(v_{0}), s(v_{1}), …, s(v_{q}) span a face of τ, and s(x) ∈ τ. We conclude that the function s: Vert K → Vert L represents a simplicial map which is a simplicial approximation to f: |K| → |L|, as required.

Corollary 4.10 If s: K → L and t: L → M are simplicial approximations to continuous maps f: |K| → |L| and g: |L| → |M|, where K, L and M are simplicial complexes, then t ∘ s: K → M is a simplicial approximation to g ∘ f: |K| → |M|.

Theorem 4.11 (Simplicial Approximation Theorem) Let K and L be simplicial complexes, and let f: |K| → |L| be a continuous map. Then, for some sufficiently large integer j, there exists a simplicial approximation s: K^{(j)} → L to f defined on the jth barycentric subdivision K^{(j)} of K.

Proof The collection consisting of the stars st_{L}(w) of all vertices w of L is an open cover of |L|, since each star st_{L}(w) is open in |L| (Lemma 4.8) and the interior of any simplex of L is contained in st_{L}(w) whenever w is a vertex of that simplex. It follows from the continuity of the map f: |K| → |L| that the collection consisting of the preimages f^{−1}(st_{L}(w)) of the stars of all vertices w of L is an open cover of |K|. It then follows from the Lebesgue Lemma that there exists some δ > 0 with the property that every subset of |K| whose diameter is less than δ is mapped by f into st_{L}(w) for some vertex w of L.

Now the mesh μ(K^{(j)}) of the jth barycentric subdivision of K tends to zero as j → +∞, since

μ(K^{(j)}) ≤ (\frac{dim K}{dim K + 1})^{j} μ(K)

for all j (Lemma 4.7). Thus we can choose j such that μ(K^{(j)}) < \frac{1}{2}δ. If v is a vertex of K^{(j)} then each point of st_{K^{(j)}}(v) is within a distance \frac{1}{2}δ of v, and hence the diameter of st_{K^{(j)}}(v) is at most δ. We can therefore choose, for each vertex v of K^{(j)} a vertex s(v) of L such that f(st_{K^{(j)}}(v)) ⊂ st_{L}(s(v)). In this way we obtain a function s: Vert K^{(j)} → Vert L from the vertices of K^{(j)} to the vertices of L. It follows directly from Proposition 4.9 that this is the desired simplicial approximation to f.

4.6 The Brouwer Fixed Point Theorem

Definition Let K be a simplicial complex which is a subdivision of some n-dimensional simplex Δ. We define a Sperner labelling of the vertices of K to be a function, labelling each vertex of K with an integer between 0 and n, with the following properties:—

• for each j ∈ {0, 1, …, n}, there is exactly one vertex of Δ labelled by j,

• if a vertex v of K belongs to some face of Δ, then some vertex of that face has the same label as v.

Lemma 4.12 (Sperner’s Lemma) Let K be a simplicial complex which is a subdivision of an n-simplex Δ. Then, for any Sperner labelling of the vertices of K, the number of n-simplices of K whose vertices are labelled by 0, 1, …, n is odd.

Proof Given integers i_{0}, i_{1}, …, i_{q} between 0 and n, let N(i_{0}, i_{1}, …, i_{q}) denote the number of q-simplices of K whose vertices are labelled by i_{0}, i_{1}, …, i_{q} (where an integer occurring k times in the list labels exactly k vertices of the simplex). We must show that N(0, 1, …, n) is odd.

We prove the result by induction on the dimension n of the simplex Δ; it is clearly true when n = 0. Suppose that the result holds in dimensions less than n. For each simplex σ of K of dimension n, let p(σ) denote the number of (n − 1)-faces of σ labelled by 0, 1, …, n − 1. If σ is labelled by 0, 1, …, n then p(σ) = 1; if σ is labelled by 0, 1, …, n − 1, j, where j < n, then p(σ) = 2; in all other cases p(σ) = 0. Therefore

∑_{{σ∈K \atop dim σ=n}} p(σ) = N(0, 1, …, n) + 2 ∑_{j=0}^{n−1} N(0, 1, …, n − 1, j).

Now the definition of Sperner labellings ensures that the only (n − 1)-face of Δ containing simplices of K labelled by 0, 1, …, n − 1 is that with vertices labelled by 0, 1, …, n − 1. Thus if M is the number of (n − 1)-simplices of K labelled by 0, 1, …, n−1 that are contained in this face, then N(0, 1, …, n− 1) − M is the number of (n − 1)-simplices labelled by 0, 1, …, n − 1 that intersect the interior of Δ. It follows that

∑_{{σ∈K \atop dim σ=n}} p(σ) = M + 2(N(0, 1, …, n − 1) − M),

since any (n−1)-simplex of K that is contained in a proper face of Δ must be a face of exactly one n-simplex of K, and any (n−1)-simplex that intersects the interior of Δ must be a face of exactly two n-simplices of K. On combining these equalities, we see that N(0, 1, …, n) − M is an even integer. But the induction hypothesis ensures that Sperner’s Lemma holds in dimension n−1, and thus M is odd. It follows that N(0, 1, …, n) is odd, as required.

Proposition 4.13 Let Δ be an n-simplex with boundary ∂Δ. Then there does not exist any continuous map r: Δ → ∂Δ with the property that r(x) = x for all x ∈ ∂Δ.

Proof Suppose that such a map r: Δ → ∂Δ were to exist. It would then follow from the Simplicial Approximation Theorem (Theorem 4.11) that there would exist a simplicial approximation s: K → L to the map r, where L is the simplicial complex consisting of all of the proper faces of Δ, and K is the jth barycentric subdivision, for some sufficiently large j, of the simplicial complex consisting of the simplex Δ together with all of its faces.

If v is a vertex of K belonging to some proper face Σ of Δ then r(v) = v, and hence s(v) must be a vertex of Σ, since s: K → L is a simplicial approximation to r: Δ → ∂Δ. In particular s(v) = v for all vertices v of Δ. Thus if v ↦ m(v) is a labelling of the vertices of Δ by the integers 0, 1, …, n, then v ↦ m(s(v)) is a Sperner labelling of the vertices of K. Thus Sperner’s Lemma (Lemma 4.12) guarantees the existence of at least one n-simplex σ of K labelled by 0, 1, …, n. But then s(σ) = Δ, which is impossible, since Δ is not a simplex of L. We conclude therefore that there cannot exist any continuous map r: Δ → ∂Δ satisfying r(x) = x for all x ∈ ∂Δ.

Theorem 4.14 (Brouwer Fixed Point Theorem) A continuous map f: E^{n} → E^{n} sending a closed n-dimensional ball E^{n} into itself has at least one fixed point (i.e., there exists x ∈ E^{n} for which f(x) = x).

Proof Suppose that the map f: E^{n} → E^{n} had no fixed point. For each x ∈ E, let q(x) be the point at which the half line starting at f(x) and passing through x intersects the boundary sphere S^{n−1} of E^{n}. Then q: E^{n} → S^{n−1} would be continuous, and q(x) = x for all x ∈ S^{n−1}. But the closed ndimensional ball E^{n} is homeomorphic to an n-simplex Δ. Therefore the map q: E^{n} → S^{n−1} would correspond under some homeomorphism h: Δ → E^{n} to a continuous map r: Δ → ∂Δ mapping Δ onto its boundary ∂Δ, where h(r(y)) = q(h(y)) for all y ∈ Δ. Moreover r(y) = y for all y ∈ ∂Δ. However Proposition 4.13 shows that there does not exist any continuous map r: Δ → ∂Δ with this property. Therefore the map f must have at least one fixed point, as required.

5 An Application of Topology to Mathematical Economics

5.1 The Existence of Equilibria in an Exchange Economy

We consider an exchange economy consisting of a finite number of commodities and a finite number of households, each provided with an initial endowment of each of the commodities. The commodities are required to be infinitely divisible: this means that a household can hold an amount x of that commodity for any non-negative real number x. (Thus salt, for example, could be regarded as an ‘infinitely divisible’ quantity whereas cars cannot: it makes little sense to talk about a particular household owning 2.637 of a car, for example, though such a household may well own 2.637 kilograms of salt.) Now the households may well wish to exchange commodities with one another so as improve on their initial endowment. They might for example seek to barter commodities with one another: however this method of redistribution would not work very efficiently in a large economy. Alternatively they might attempt to set up a price mechanism to simplify the task of redistributing the commodities. Thus suppose that each commodity is assigned a given price. Then each household could sell its initial endowment to the market, receiving in return the value of its initial endowment at the given prices. The household could then purchase from the market a quantity of each commodity so as to maximize its own preference, subject to the constraint that the total value of the commodities purchased by any household cannot exceed the value of its initial endowment at the given prices. The problem of redistribution then becomes one of fixing prices so that there is exactly enough of each commodity to go around: if the price of any commodity is too low then the demand for that commodity is likely to outstrip supply, whereas if the price is too high then supply will exceed demand. A Walras equilibrium is achieved if prices can be found so that the supply of each commodity matches its demand. We shall use the Brouwer fixed point theorem to prove the existence of a Walras equilibrium in this idealized economy.

Let our exchange economy consist of n commodities and m households. We suppose that household h is provided with an initial endowment \overline{x}_{hi} of commodity i, where \overline{x}_{hi} ≥ 0. Thus the initial endowment of household h can be represented by a vector \overline{x}_{h} in ℝ^{n} whose ith component is \overline{x}_{hi}. The prices of the commodities are given by a price vector p whose ith component p_{i} specifies the price of a unit of the ith commodity: a price vector p is required to satisfy p_{i} ≥ 0 for all i. Then the value of the initial endowment of household h at the given prices is p.\overline{x}_{h}. Let x_{hi}(p) be the quantity of commodity i that household h seeks to purchase at prices p, and let x_{h}(p) ∈ ℝ^{n} be the vector whose ith component is x_{hi}(p). The budget constraint certainly ensures that p.(x_{h}(p) − \overline{x}_{h}) ≤ 0 (i.e., the value of the goods purchased cannot exceed the value of the initial endowment at the given prices). We assume that the value of the commodities that each household seeks to purchase is equal to the value of its initial endowment, and thus p.x_{h}(p) = p.\overline{x}_{h}. Also the preferences of the household will only depend on the relative prices of the commodities, and therefore x_{h}(λp) = x_{h}(p) for all λ > 0.

Now the total supply of each commodity in the economy is represented by the vector ∑_{h} \overline{x}_{h}, and the total demand at prices p is represented by ∑_{h} x_{h}(p). The excess demand in the economy at prices p is therefore represented by the vector z(p), where z(p) = ∑_{h}(x_{h}(p) − \overline{x}_{h}). Let z_{i}(p) be the ith component of z(p). Then z_{i}(p) > 0 when the demand for the ith commodity exceeds supply, whereas z_{i}(p) < 0 when the supply exceeds demand. Note that p.z(p) = 0 for any price vector p. This identity, known as Walras’ Law, follows immediately on summing the budget constraint p.x_{h}(p) = p.\overline{x}_{h} over all households.

Theorem 5.1 Consider an exchange economy consisting of a finite number of infinitely divisible commodities and a finite number of households. Let the excess demand in the economy at prices p be given by z(p), where

(i) the excess demand vector z(p) is well-defined for any price vector p, and depends continuously on p,

(ii) p.z(p) = 0 for any price vector p (Walras’ Law).

(iii) Then there exist equilibrium prices p^{∗} at which z_{i}(p^{∗}) ≤ 0 for all i.

Proof Let Δ be the (n−1)-dimensional simplex in ℝ^{n} consisting of all points (p_{1}, p_{2}, …, p_{n}) in ℝ^{n} satisfying 0 ≤ p_{i} ≤ 1 for i = 1, 2, …, n and ∑_{i=1}^{n} p_{i} = 1, and let v: Δ → ℝ^{n} be the function with ith component v_{i} given by

v_{i}(p) = \{ \begin{matrix}p_{i} + z_{i}(p) & if z_{i}(p) > 0; \\ p_{i} & if z_{i}(p) ≤ 0. \end{matrix}

Note that v(p) ≠ 0 and the components of v(p) are non-negative for all p ∈ Δ. It follows that there is a well-defined map φ: Δ → Δ given by

φ(p) = \frac{1}{∑_{i=1}^{n} v_{i}(p)}v(p),

The Brouwer Fixed Point Theorem (Theorem 4.14) ensures that there exists p^{∗} ∈ Δ satisfying φ(p^{∗}) = p^{∗}. Then v(p^{∗}) = λp^{∗} for some λ ≥ 1. We claim that λ = 1.

Suppose that it were the case that λ > 1. Then v_{i}(p^{∗}) > p_{i}^{∗}, and thus z_{i}(p^{∗}) > 0 whenever p_{i}^{∗} > 0. But p_{i}^{∗} ≥ 0 for all i, and p_{i}^{∗} > 0 for at least one value of i, since p^{∗} ∈ Δ. It would follow that p^{∗}.z(p^{∗}) > 0, contradicting Walras’ Law. We conclude that λ = 1, and thus v_{i} = p_{i}^{∗} and z_{i}(p^{∗}) ≤ 0 for all i, as required.

Note that if z_{i}(p^{∗}) ≤ 0 for all i and p^{∗}.z(p^{∗}) = 0, then p_{i}^{∗}z_{i}(p^{∗}) ≤ 0 for all i, and z_{i}(p^{∗}) = 0 whenever p_{i} > 0. Thus, at equilibrium prices, supply always equals or exceeds demand, and supply equals demand for those commodities with positive prices.

The proof of the existence of Walras equilibria can readily be generalized to Arrow-Debreu models where economic activity is carried out by both households and firms. The problem of existence of equilibria was studied by L. Walras in the 1870s, though a rigorous proof of the existence of equilibria was not found till the 1930s, when A. Wald proved existence for a limited range of economic models. Proofs of existence using the Brouwer Fixed Point Theorem, or a more general fixed point theorem due to Katukani, were first published in 1954 by K. J. Arrow and G. Debreu and by L. McKenzie. Subsequent research has centred on problems of uniqueness and stability, and the existence theorems have been generalized to economies with an infinite number of commodities and economic agents (households and firms). An alternative approach to the existence theorems using techniques of differential topology was pioneered by G. Debreu and by S. Smale.

More detailed accounts of the theory of ‘general equilibrium’ can be found in, for example, The theory of value, by G. Debreu, General competitive analysis, by K. J. Arrow and F. H. Hahn, or Economics for mathematicians by J. W. S. Cassels.