Introduction to Topology

Homotopy Theory

THREE

Many questions that arise naturally in topology are difficult to handle if only the basic definitions and their immediate consequences are used. It is certainly natural to inquire whether two given topological spaces are homeomorphic, and yet this question is hard to answer even for familiar spaces. Intuition suggests that the punctured plane ²\{(0,0)} is not homeomorphic to the plane ², but a precise proof of this fact using only the foundational material of Chapter II is difficult. A successful approach to such problems involves a principle that at first description may seem unduly abstract. It is the principle of associating with topological objects certain algebraic objects, thus converting topological problems into algebraic problems, which, one hopes, are more tractable. This idea has developed into a whole branch of mathematics, algebraic topology. The purpose of this chapter is to develop an important concrete instance of the general process.

We begin in Section 1 by introducing the appropriate algebraic concept, that of a group. Sections 2 through 6 form a unit in which the fundamental group of a topological space is introduced and the fundamental group of the circle is computed from a geometric point of view using covering spaces. Several elementary applications are given, such as the Brouwer Fixed Point Theorem in dimension two. In Sections 7 and 8 we discuss homotopic maps and give an alternative computation of the fundamental group of the circle, this time from an analytic point of view. Sections 9 and 10 include deeper applications of the ideas introduced in this chapter. In Section 9 we prove that every vector field on the 2-sphere has a zero, and in Section 10 we prove the deceptively difficult Jordan Curve Theorem, which states that every simple closed curve in the plane has an “inside” and an “outside.”

1. GROUPS

A group is a set G, and a mapping (a,b) → a * b of G × G into G, that satisfy the following axioms:

Note that the identity of a group is unique. Indeed, if e and e′ both satisfy (1.2), then e = e _* e′ = e′.

If both b and b′ are elements of G that satisfy (1.3), then

b = b _* e = b _* (a _* b′) = (b _* a) _* b′ = e _* b′ = b′.

Consequently the element b satisfying (1.3) is unique. It is called the inverse of a and denoted by a^–1, so that

a _* a⁻¹ = a⁻¹ _* a = e, a G.

By (1.2), e^–1 = e.

The group G is commutative, or abelian, if a _* b = b _* a for all a, b G.

Examples of groups abound. Here is a list of a few common groups:

(i) The trivial group consisting of only one element, the identity, with the only possible operation, is obviously a commutative group.

(ii) The set of integers, with the operation of addition, forms an abelian group. The integer 0 is the identity of , and the inverse of m is – m.

(iii) The set of real numbers and the set of complex numbers, with the operation of addition, are commutative groups.

(iv) Any real or complex vector space, with the operation of addition, is a commutative group. In particular, ⁿ and ⁿ are commutative groups.

(v) The set of nonzero real numbers, with the operation of multiplication, is a commutative group. The identity is 1, and the inverse of s is 1/s.

(vi) The unit circle S¹ in the complex plane, with the operation of multiplication, is a commutative group. The identity is 1, and the inverse of a S¹ is the complex conjugate ā of a.

(vii) The set GL(n, ) of invertible n × n matrices with real entries, with matrix multiplication as the operation, forms a group. If we allow complex entries, we obtain a group GL(n, ). These groups are called general linear groups. For n ≥ 2, they are not commutative.

(viii) The set S_n of permutations of n objects, with composition as the operation, forms a group called the symmetric group. The group S_n has n! elements. If n ≥ 3, S_n is not commutative.

(ix) The set _n of congruence classes of integers modulo n, with the operation of addition, forms a commutative group of n elements.

A subset H of a group G is a subgroup of G if H forms a group when endowed with the multiplication inherited from G. In order that a subset H of G be a subgroup of G, it is necessary and sufficient that H be “closed” under multiplication and taking inverses, that is, that

For example, is a subgroup of and is a subgroup of (operation of addition). Subspaces of a vector space are in particular subgroups of the vector space. However, a subgroup of a vector space need not be a subspace, as indicated by the example ⊂ .

Now let G and H be two groups. A function f from G to H is a homomorphism if

Here the product a _* b is taken in G and the product f(a) _* f(b) is taken in H. For instance, a map h from (operation of addition) to the nonzero real numbers (operation of multiplication) is a homomorphism if and only if

h(s + t) = h(s)h(t), s, t .

An example of such a map is given by

h(t) = e^t, t .

Some other examples of homomorphisms are as follows:

(i) The map f : → defined by

f(m) = 5m, m .

(ii) The map f : → S¹ defined by

f(s) = e^is, s .

(iii) The map f : → _n defined so that f(m) is the congruence class of m (modulo n).

(iv) The map f : → GL(2,) defined by

(v) If H is a subgroup of G, then the inclusion map is a homomorphism.

Let G and H be groups and let f : G → H be a homomorphism. If e is the identity of G and e′ is the identity of H, then f(e) = e′. This follows from the chain of identities

e′ = f(e) _* f(e)⁻¹ = f(e _* e) _* f(e)⁻¹ = f(e) _* f(e) _* f(e)⁻¹ = f(e).

(Here we ignore the order in which the operations are performed since they are associative.) From the identity e′ = f(e) = f(aa⁻¹) = f(a)f(a⁻¹), we conclude that

f(a⁻¹) = f(a)⁻¹, a G.

If K is another group and both f : G → H and g : H → K are homomorphisms, then the composition g ∘ f : G → K is also a homomorphism. Indeed,

whenever a, b G.

A homomorphism f : G → H is an isomorphism if f is one-to-one and onto. If there is an isomorphism of G and H, then G and H are said to be isomorphic, written G ≅ H. The composition of two isomorphisms is an isomorphism. Furthermore, the inverse of an isomorphism is an isomorphism. Indeed, if x, y H, say x = f(a) and y = f(b), then f(a _* b) = x _* y, so that

f⁻¹(x _* y) = a _* b = f⁻¹(x) _* f⁻¹(y).

One familiar isomorphism is the map f : ² → given by

f((x,y)) = x + iy, (x,y) ².

Another example of an isomorphism is the function f from to the group ₊ of positive real numbers, with the operation of multiplication, given by

f(t) = e^t, t .

EXERCISES

1. Prove that any two groups with one element are isomorphic. Prove that any two groups with two elements are isomorphic. Prove that any two groups with three elements are isomorphic.

2. Show that for any fixed positive integer n, there are at most a finite number of nonisomorphic groups of order n. Hint: If α₁, . . . ,α_n are the elements of the group, then the group is determined by knowing all the products α_i _* α_j, 1 ≤ i ≤ j ≤ n.

3. Find the number of nonisomorphic groups of order 4.

4. Prove that if f is a homomorphism from a group G to a group H, then f(G) is a subgroup of H. Prove that if e is the identity of H, then f^–1(e) is a subgroup of G.

5. Let G and H be groups. Define an operation in the Cartesian product G × H by

(a₁, b₁) _* (a₂, b₂) = (a₁, _* a₂, b₁ _* b₂), a₁, a₂ G; b₁, b₁, b₂ H.

Show that endowed with this operation, G × H becomes a group. Show that G × H is commutative if and only if G and H are commutative.

Remark. G × H is the direct product of G and H. If G and H are abelian and if the operations in G and H are denoted by “ + ”, then it is customary to refer to the direct product of G and H as the direct sum of G and H and denote it by G ⊕ H. For example, the group ⊕ is isomorphic to ² and ⊕ is isomorphic to the subgroup of integral lattice points in ².

2. HOMOTOPIC PATHS

Let X be a topological space and let a, b X. Recall that a path in X from a to b is a map γ : [0,1] → X such that γ(0) = a and γ(1) = b. Two paths γ₀ and γ₁ from a to b are homotopic with endpoints fixed, written γ₀ γ₁ rel{0,1}, if there is a map

F : [0,1] × [0,1] → X

such that

The map F is referred to as a homotopy of γ₀ and γ₁. For each t [0,1], the map γ_t : [0,1] → X defined by

is then a path from a to b. The variable t is to be regarded as a parameter, and the paths γ_t move “continuously” with t. As the parameter t moves from 0 to 1, the path γ₀ is continuously deformed to the path γ₁ through the γ_t’s. Often we shall refer to {γt}_0≤t≤1 as a homotopy of γ₀ and γ₁, understanding that the homotopy F can be recovered from the γ_t’s by (2.1). The situation may be represented by a square, representing the parameter space, with left edge labeled “a” to indicate that the restriction of the homotopy F to the left edge is the constant path at a, with bottom edge labeled “γ₀” to indicate that the restriction of F to the bottom edge coincides with γ₀, etc. (see the diagram).

A subset X of ⁿ is convex if, whenever x and y belong to X, then the straight-line interval joining x to y lies in X. In other words, X is convex if, whenever x, y X, then tx + (1 – t)y X for 0 ≤ t ≤ 1. The following theorem provides the fundamental example of homotopic paths.

2.1 Theorem: Let X be a convex subset of ⁿ and let a, b X. Then any two paths in X from a to b are homotopic in X with endpoints fixed.

Proof: Let γ₀ and γ₁ be paths in X from a to b. The homotopy {γ_t} is defined so that for each fixed s, γ_t(s) moves from γ₀(s) to γ₁(s) along the straight-line interval joining the points. An explicit homotopy is given by defining

F(s,t) = γ_t(s) = (1 − t)γ₀(s) + tγ₁(s), 0≤ s, t ≤ 1.

One checks that each γ_t is a path in X from a to b and that the γ_t move continuously with t, that is, the homotopy F is continuous.

The remainder of this section is devoted to a series of elementary lemmas, which will form the foundation for the discussion of the fundamental group in Section 3. We fix a topological space X.

2.2 Lemma: The relation γ α rel{0,1} is an equivalence relation on the set of paths in X from a to b.

Proof: A homotopy of γ to γ is given by γ_t = γ, 0 ≤ t ≤ 1, so that the relation is reflexive.

Suppose that γ₀ γ₁ rel{0,1} and let {γ_t} be a homotopy of γ₀ and γ₁. Then by changing the direction of the parameter, we obtain a homotopy t → γ_{1 – t}, 0 ≤ t ≤ 1, of γ₁ to γ₀. In other words,

F(s,t) = γ_{1 − t}(s), 0 ≤ s, t ≤ 1,

is a homotopy of γ₁ and γ₀. Hence the relation is symmetric.

Suppose that γ₀ γ₁ rel{0,1} and γ₁ γ₂ rel{0,1}. Let {γ_t}_0≤t≤1 be the homotopy of γ₀ and γ₀ and let {α_t}_0≤t≤1 be the homotopy of γ₁ to γ₂, so that α₀ = γ₁ and α₁ = γ₂. Set

Since γ₁ = α₀, β_1/2 is well defined. The homotopy β_t corresponds to deforming γ₀ to γ₁ at double speed and then deforming γ₁ to γ₂ at double speed. The map (s,t) → β_t(s) is continuous, so that {β_t} is indeed a homotopy of γ₀ and γ₂, and the relation is transitive.

The equivalence classes of paths in X from a to b modulo this equivalence relation are called the homotopy classes of paths from a to b. The homotopy class of a path γ is denoted by [γ]. Then [γ₀] = [γ₀] means that γ₀ γ₁ rel{0,1}, that is, that γ₀ and γ₁ are homotopic with endpoints fixed.

The next lemma shows that any reparametrization of a path lies in the same homotopy class.

2.3 Lemma: Let γ be a path in X from a to b. Let ρ be any map from [0,1] to [0,1] such that ρ(0) = 0 and ρ(1) = 1. Then [γ∘ρ] = [γ].

Proof: Note that γ∘ρ is a path from a to b. It is obtained by running along γ at a varying speed, with perhaps some backsliding. A homotopy of γ to γ∘ρ is given explicitly by

γ_t(s) = γ((1 − t)s + tρ(s)), 0 ≤ s, t ≤ 1.

This homotopy is precisely the composition of γ and the homotopy α_t(s) = (1 − t)s + tρ(s) of ρ, regarded as a path in [0,1] from 0 to 1, and the identity path α₀(s) = s from 0 to 1. In particular, γ_t is well defined and moves continuously with t.

Now let a, b, c X, let α be a path from a to b, and let β be a path from b to c. As in Section II.9, we define a path αβ from a to c by

In other words, αβ is obtained by running first along α at double speed and then along β at double speed.

2.4 Lemma: Let a, b, c X, let α₀ and α₁ be paths from a to b, and let β₀ and β₁ be paths from b to c. If [α₀] = [α₁] and [β₀] = [β₁] then [α₀β₀] = [α₁β₁].

Proof: If {α_t}_0≤t≤1 is a homotopy of α₀ and α₁ and {β_t}_0≤t≤1 is a homotopy of β_O and β₁, then {α_tβ_t}_{0≤t ≤1} is a homotopy of α₀β₀ and α₁β₁.

Lemma 2.4 allows us to define under certain circumstances the products of homotopy classes. If α is a path from a to b and β is a path from b to c, we define

[α][β] = [αβ]

By Lemma 2.4, the class [αβ] does not depend on the representatives of [α] and [β] used to define it.

Consider three paths α, β, and γ such that the terminal point of α coincides with the initial point of β and the terminal point of β coincides with the initial point of γ. Then (αβ)γ and α(βγ) are both defined, and they are given by

and

It is easy to give examples of α, β, and γ such that (αβ)γ ≠ α(βγ). In other words, multiplication of paths (when defined) is not an associative operation. However, when we pass to homotopy classes, we do obtain an associative operation.

2.5 Lemma: Let α, β, and γ be paths in X as above, so that (αβ)γ and α(βγ) are defined. Then

(αβ)γ α(βγ) rel{0,1}.

In other words,

([α][β][γ]) = [α]([β][γ]).

Proof: Comparing the formulas above for the respective products, we find that

(α(βγ))(s) = ((αβ)γ)(ρ(s)), 0 ≤ s ≤ 1,

where

In other words, α(βγ) is a reparametrization of (αβ)γ, so that, by Lemma 2.3, the paths are homotopic with endpoints fixed.

The constant path at the point b will also be denoted by b. It is defined by b(s) = b, 0 ≤ s ≤ 1.

2.6 Lemma: If α is a path from a to b, then

aα α αb rel{0,1}.

In other words,

[a][α] = [α] = [α][b].

Proof: The path aα is given by

then aα = α∘ρ. By Lemma 2.3, aα α rel{0,1}. Similarly, αb α rel{0,1}.

Now let a be a path in X from a to b. As in Section II.9, we define a path α^–1 from b to a by

(α^–1)(s) = α(1 – s), 0 ≤ s ≤ 1,

so that α^–1 corresponds to running backwards along α. In particular, for the constant path b, we have b⁻¹ = b.

2.7 Lemma: Let α₀ and α₁ be paths in X from a to b. If [α₀] = [α₁] then [α₀^–1] = [α₁^–1].

Proof: If {α_t}_0≤t≤1 is a homotopy of α₀ and α₁, then {α_t^{− 1}}_0≤t≤1 is a homotopy of α₁⁻¹ and α₀⁻¹.

Lemma 2.7 allows us to define the inverse of the homotopy class of a path α by

[α]⁻¹ = [α⁻¹].

By the lemma, the homotopy class defining [α]^–1 does not depend on the choice of the representative of the homotopy class of α.

The final preparatory lemma asserts that the path obtained by running along α at double speed and then returning backwards along α at double speed is homotopic to a constant path. Note that in a homotopy of a product path, the intermediate stopping points are not required to be held fixed.

2.8 Lemma: Let α be a path in X from a to b. Then αα⁻¹ is homotopic to the constant path at a with endpoints fixed. In other words,

[α][α]⁻¹ = [a]

Proof: For 0 ≤ t ≤ 1, let γ_t be the path described by running along α at double speed until time t/2, then resting until time 1 − t/2, then returning along α at double speed. The explicit formula for γ_t is

Then γ₀ is the constant path at a and γ₁ and αα⁻¹. It is easy to check that the γ_t move continuously with t, so that they form a homotopy.

EXERCISES

1. Suppose that (αβ)γ = α(βγ) for any three paths α, β, and γ in X for which the product is defined. Show that each path comaponent of X consists of a single point.

2. Let X be path-connected and let b X. Show that every path in X is homotopic with endpoints fixed to a path passing through b.

3. Let D be an open subset in ⁿ, let α be a path in D from x to y, and set

d = inf{|α(s) − w| : w ∂D, 0 ≤ s ≤ 1}.

Show that if β is any path in D from x to y such that |β(s) − α(s)| < d, 0 ≤ s ≤ 1, then β is homotopic to α with endpoints fixed.

4. A path a in ⁿ, is polygonal if there is a subdivision 0 = s₀ < s₁ < · · · < s_m = 1 of the unit interval such that a maps each interval [s_{j − 1}, s_j] onto the straight-line segment from to α(s_{j – 1}) to α(s_j). Show that every path in an open subset D of ⁿ, is homotopic in D with endpoints fixed to a polygonal path.

5. Show that any path in Sⁿ is homotopic with endpoints fixed to a polygonal path on Sⁿ, where “polygonal” is now interpreted to mean that the path is formed from arcs lying on great circles of Sⁿ.

6. Let (X,d) be a compact metric space and let a, b X. Let be the set of all paths in X from a to b with the metric

ρ(α, β) = sup{d(α(s), β(s)) : 0 ≤ s ≤ 1}.

Show that two paths α,β are homotopic with endpoints fixed if and only if α and β lie in the same path component of .

3. THE FUNDAMENTAL GROUP

A pointed space is a pair (X,b) consisting of a topological space X and a point b X. The point b is referred to as the base point of X. A loop in X based at b is a path in X that begins and ends at b. The product of any two loops based at b is well defined, so that the product of the homotopy classes of any two such loops is well defined. Let π₁(X,b) denote the set of homotopy classes of loops based at b, together with the multiplication of homotopy classes defined in Section 2.

3.1 Theorem: The set π₁(X,b) of homotopy classes of loops based at b, with the operation of multiplication of homotopy classes, is a group.

Proof: Lemma 2.5 shows that the multiplication in π₁(X,b) is associative. Lemma 2.6 shows that the homotopy class [b ] of the constant map b is an identity for π₁(X,b) Lemma 2.8 shows that the homotopy class [α]⁻¹ of α⁻¹ is an inverse for the homotopy class of α.

The group π₁(X,b) is called the fundamental group of X based at b. If π₁(X,b) consists of only the identity, we say that π₁(X,b) is trivial and write π₁(X,b) ≅ 0. Thus π₁(X,b) is trivial if and only if every loop based at b is homotopic with endpoints fixed to the constant loop at b. Theorem 2.1 then yields the following result.

3.2 Theorem: If X s a convex subset of ⁿ, and b X, then π₁(X,b) ≅ 0.

Reclining figure eight

It turns out that the fundamental group of a space need not be commutative. For instance, if X is a “figure eight,” then π₁(X,b) is isomorphic to a certain non-commutative group, namely, the free group on two generators. This theorem is due to Van Kampen, and a proof will be indicated in the exercises in Section 5. Meanwhile, it will take us some effort to prove that there is a space X such that π₁(X,b) is not trivial. The simplest space with a nontrivial fundamental group is the circle S¹, and it will be shown in Section 5 that

π₁(S¹, 1) ≅ .

Now the question arises of how π₁(X,b) depends on the base point b. Any loop based at b lies within the path component of b in X. Therefore if c X lies in a path component different from b, then π₁(X,c) is in no way related to π₁(X,b). However, if c lies in the same path component as b, then π₁(X,c) is isomorphic to π₁(X,b). In fact, the following is true.

3.3 Theorem: Let b, c X and suppose α to be a path in X from b to c. Then for each loop γ based at c, the homotopy class

α_*([γ]) = [α][γ][α]⁻¹

is defined and α_* is an isomorphism of π₁(X,c) and π₁(X,b).

Proof: Since the initial and terminal points of the paths match up appropriately, α_* is well defined. Since the multiplication of homotopy classes is associative, the parentheses indicating the order of multiplication are omitted.

Suppose that β and γ are both loops at c. From Lemmas 2.6 and 2.8 we obtain

[β][γ] = [β][c][γ] = [β][α]⁻¹[α][γ].

Consequently

Hence α_* is a homomorphism.

Suppose next that the loops β and γ at c satisfy α_*([β]) = α_*([γ]). Then

[β] = [α]⁻¹[α][β][α]⁻¹[α] = [α]⁻¹[α][γ][α]⁻¹[α] = [γ].

Hence α_* is one-to-one.

Finally, let λ be a loop based at b. Then β = (α⁻¹λ)α is a loop based at c that satisfies

α_*([β]) = [α][α]⁻¹[λ][α][α]⁻¹ = [λ].

Consequently α_* is onto and is an isomorphism.

If X is path-connected, then the various groups π₁(X,b), b X, are all isomorphic. Thus the isomorphism type of π₁(X,b) is well defined; it is called the fundamental group of X, denoted by π₁(X). The situation here is slightly peculiar linguistically in that the fundamental group π₁(X) is not equal to any of the π₁(X,b) but is rather the (common) isomorphism class of all the distinct but isomorphic groups π₁(X,b), b X. In the following development, the assertions to be made about π₁(X) will be such assertions as “π₁(X) is isomorphic to G,” where G is some concretely described group. This assertion has the precise meaning that, for all b X, π₁(X,b) is isomorphic to G; the isomorphism holds for all b in X if it holds for any one b in X since all π₁(X,b) are isomorphic.

A space X is simply connected if X is path-connected and π₁(X) is trivial. Since any convex subset of ⁿ is path-connected, Theorem 3.2 shows that a convex subset of ⁿ is simply connected.

There is another way of viewing loops in X based at b, which is convenient for some purposes. Let S¹ denote the unit circle in ² ≅ . Then each map

f : S¹ → X

satisfying f(1) = b determines a loop α based at b via the formula

α(s) = f(e^2πis), 0 ≤ S ≤ 1.

Conversely, any loop α based at b arises from a map f : S¹ → X satisfying f(1) = b. The point is that the exponential map s → e^2πis is a homeomorphism from the interval [0,1], with the endpoints identified, to the circle S¹.

If the loops α₀ and α₁ based at b are homotopic with endpoints fixed, then the corresponding maps f₀, f₁ : S¹ → X are homotopic relative to the base points 1 S¹ and b X in the sense that there are maps {f_t}_0≤t≤1 from S¹ to X such that f_t(1) = b, 0 ≤ t ≤ 1, and the f_t move continuously with t. In other words, there is a map

F : S¹ × [0,1] → X

such that

Conversely, if f₀ and f₁ are homotopic as above, then (s,t) → F(e^2πis, t) is a homotopy of α₀ and α₁ with endpoints fixed.

With this identification in mind, we shall think of maps f : (S¹, 1) → (X,b) also as loops in X based at f(1).

EXERCISES

1. Prove that if n ≥ 2, then Sⁿ is simply connected. Hint: Use Exercise 2.5 to show that every loop in Sⁿ is homotopic to a loop that does not cover all of Sⁿ.

2. Prove that if there are simply connected open subsets U and V of X such that U ∪ V = X and U ∩ V is nonempty and path-connected, then X is simply connected.

3. A space X is contractible to a point x₀ X with x₀ held fixed if there is a map F : X × [0, 1] → X such that

Show that such a space is simply connected.

4. Let X be the comb space, that is, the compact subset of ² consisting of the horizontal interval {(x, 0) : 0 ≤ x ≤ 1} and the closed vertical intervals of unit length with lower endpoints at (0,0) and at (0,1/n), 1 ≤ n < ∞.

Comb space

Show that X is contractible to (0,0) with (0,0) held fixed. Show that X is not contractible to (0,1) with (0,1) held fixed.

5. A subset W of ⁿ is star-shaped with respect to a point w W if, whenever y W, then the straight-line segment from w to y is contained in W.

(a) Show that a subset W of ⁿ is convex if and only if it is star-shaped with respect to each of its points.

(b) Give an example of a star-shaped set that is not convex.

(d) Show that a star-shaped set in ⁿ is simply connected.

6. Let (X,x₀) and (Y,y₀) be pointed spaces. Show that π₁(X × Y, (x₀y₀)) is isomorphic to the direct product π₁(X,x₀) × π₁(Y,y₀).

7. Prove that the product of simply connected spaces is simply connected.

8. Prove that if n ≥ 3, then ⁿ\{0} is simply connected.

9. Let X be a path-connected topological space and let b, c X. Let B² be the closed unit ball in ², with boundary circle S¹. Show that the following are equivalent.

(a) X is simply connected.

(b) Any two paths from b to c are homotopic with endpoints fixed.

4. INDUCED HOMOMORPHISMS

Let (X,b) and (Y,c) be pointed topological spaces. A map f : (Y,c) → (X,b) is a continuous function f from Y to X that satisfies f(c) = b. We aim to show that any such map induces a homomorphism f_* from π₁(Y,c) to π₁(X,b). Despite the notational similarity, f_* is unrelated to the induced map α_* of the preceding section. The map f_* will be defined in the obvious way, by making the path γ in Y correspond to the path f∘γ in X. That this correspondence respects homotopy classes is the content of the following elementary but useful lemma.

4.1 Lemma: Let X and Y be topological spaces and let f : Y → X be a map. Let α₀ and α₁ be paths in Y that are homotopic with endpoints fixed. Then f∘α₀ and f∘α₁ are paths in X that are homotopic with endpoints fixed.

Proof: If {α_t}_0≤t≤1 is the homotopy of α₀ to α₁, then the paths {f_∘αt}_0≤t≤1 form a homotopy of f∘α₀ and f∘α₁.

The following corollary to Lemma 4.1 and Theorem 2.1 is sufficiently useful to merit a separate statement.

4.2 Corollary: Let Y be a convex subset of ⁿ let y₀, y₁ Y, and let f be a map from Y to X. If α and β are paths in Y from y₀ to y₁, then f∘α is homotopic to f∘β with endpoints fixed.

Suppose now that c₀, c₁ Y and that f is a map from Y to X. If α is a path in Y from c₀ to c₁, then f_*([α]) is defined to be the homotopy class of paths in X from f(c₀) to f(c₁) that includes f∘α:

f_*([α]) = [f∘α].

By Lemma 4.1, this definition does not depend on the choice of the representative of the homotopy class [α]. If the path product αβ is defined, then f∘(αβ) = (f∘α)(f∘β), so that

Since the inverse of the path f∘α is f∘(α⁻¹),

f_*([α]⁻¹) = f_*([α])⁻¹.

Finally,

f_*([c] = [f(c)]

whenever c is a constant path in Y.

4.3 Theorem: Let (X,b) and (Y,c) be pointed topological spaces and let f : (Y,c) → (X,b) be a map. Then f_* is a homomorphism of π₁(Y,c) and π₁(X,b). If, furthermore, (W,d) is a pointed topological space and g : (W,d) → (Y,c) is a map, then

(f∘g)_* = f_*∘g_*.

Finally, if X = Y, b = c, and f is the identity map, then f_* is the identity isomorphism.

Proof: That f_* is a htomomorphism follows from (4.1). The other statements follow directly from the definitions.

4.4 Corollary: If f : Y → X is a homeomorphism and if c Y and b = f(c), then f_* is an isomorphism of π₁(Y,c) and π₁(X,b).

Proof: Since f∘f⁻¹ and f⁻¹∘f are the identity maps of X and Y, respectively, f_*∘(f⁻¹)_* and (f⁻¹)_*∘f_* are the identity isomorphisms of π₁(X,b) and π₁(Y,c), respectively. Since (f⁻¹)_* ∘f_* is one-to-one, so is f_*. Since f_* ∘(f⁻¹) _* is onto, so is f_*. Hence f_* is an isomorphism.

With Corollary 4.4, we have attained the goal indicated at the beginning of the chapter, namely, we have associated to each (pointed) topological space an algebraic object, its fundamental group, in such a way that if two spaces are homeomorphic (via a base-point-preserving homeomorphism), then the algebraic objects are isomorphic. Thus, for instance, two spaces could be shown to be nonhomeomorphic by showing that their fundamental groups were not isomorphic. To give this idea any real significance, it is necessary to be able to compute the fundamental groups of whatever topological spaces one wishes to study. In the following section, we shall show how to determine the fundamental group of the circle S¹—it turns out to be (isomorphic to) the group of integers with addition as the group operation. We focus attention first on the circle S¹ because it is obviously the simplest space containing a closed curve without self-intersections. The method used to compute π₁(S¹) will serve to compute the fundamental group of many other topological spaces. Some applications of the fundamental group to fixed-point theorems and other related results will be presented in Section 6 and later sections.

EXERCISES

1. Show that simple connectivity is a topological property.

2. A subspace A of a topological space X is a retract of X if there is a map f : X → A such that f(y) = y for all y A. The map f is called a retraction of X onto A. Show that the unit sphere Sⁿ in ^{n + 1} is a retract of ^{n + 1}\{0}.

3. Let f be a retraction of X onto A and let x₀ A. Let j : be the inclusion map. Prove the following:

(a) j_* : π₁(A,x₀) → π₁(X,x₀) is one-to-one.

(b) f_* : π₁(X,x₀)→ π₁(A,x₀) is onto.

5. COVERING SPACES

We wish now to compute the fundamental group of the circle S¹. This will be accomplished with the aid of the exponential map p : → S¹, defined by

Since the line of proof will be quite general, we axiomatize those properties of the spaces and S¹ and the exponential map p that will be needed.

Let E and X be topological spaces and let p : E → X be a map. An open subset U of X is evenly covered by p if the inverse image p⁻¹(U) is a union of disjoint open subsets of E, each of which is mapped homeomorphically by p onto U. The map p is a covering map if p maps E onto X and if each x X has an open neighborhood that is evenly covered by p. In this case, E is a covering space over X.

If x X, the set p⁻¹(x) is called the fiber over x. It is evidently a discrete subspace of E. According to the definition, each x X has an open neighborhood U such that p⁻¹(U) is homeomorphic to p⁻¹(x) × U. The subsets of p⁻¹(U) that are mapped homeomorphically onto U are called the sheets of p⁻¹(U). If U is connected, then the sheets of p⁻¹(U) coincide with the connected components of p⁻¹(U).

The exponential map p : → S¹ is a covering map. Indeed, let e^2πit₀ S¹, fix 0 < ε < 1/2, and let U = {e^2πit : |t − t₀| < ε}. Then p⁻¹(U) is the disjoint union of the intervals (m + t₀ − ε, m + t₀ + ε), m an integer, and each of these intervals is mapped homeomorphically by p onto U. For the purposes of visualizing the covering map, it is convenient to think of as the helix {(cor(2πt), sin(2πt), t) : − ∞ < t < ∞} in ³ by identifying t with the corresponding point on the helix. The covering map p then projects the helix onto the circle S¹ in the x, y-plane.

As a second example, let X be a figure eight, consisting of two loops touching at one point b. Let E₀ be the space obtained by unwinding one of the loops, so that E₀ is a helix with loops attached as indicated in the figure above. If we unwind the loop of E₀ touching the base point e₀ of E₀, we obtain another covering space E over X, as suggested by the figure appearing towards the end of this section. The covering map is the composition of two covering maps, one from E to E₀, the other from E₀ to X.

As another example, consider the n-dimensional projective space Pⁿ, defined in Exercise II.13.8. Recall that Pⁿ is the quotient space obtained from the unit sphere Sⁿ in ^{n + 1} by identifying antipodal points. Let p : Sⁿ → Pⁿ be the quotient map. If x₀ Sⁿ, > 0 is small, and V = {x Sⁿ : |x − x₀| < ε}, then U = p(V) is an open subset of Pⁿ and p⁻¹(U) is the disjoint union of V and −V. Since both V and −V are mapped homeomorphically by p onto U, p is a covering map. In this case, every fiber consists of two points.

Products of covering spaces are covering spaces. For instance, if Tⁿ = S¹ × . . . × S¹ is the n-dimensional torus (product of n circles), then the exponential map p : ⁿ → Tⁿ, defined by

p(x₁, . . . , x_n) = (e^2πix₁, . . . , e^2πix_n),

is a covering map.

Now let p : E → X be a covering map, let Y be a topological space, and let f : Y → X be a map. For later developments it will be important to determine whether there is a map g : Y → E such that p∘g = f. Such a map g is called a lift of f. The situation may be represented schematically by the diagram

The relation p∘g = f means that the image of a point y Y is the same whether f is applied directly (the “low road”) or g is applied followed by the projection p (the “high road”). If p∘g = f, the diagram is said to commute.

The uniqueness of lifts will be handled by the following lemma.

5.1 Lemma: Let p : E → X be a covering map and let Y be a connected topological space. Let f : Y → X be a map and let g, h : Y → E be two lifts of f. If g(y) = h(y) for some point y Y, then g = h.

Proof: Let

Then S ∪ T = Y and S ∩ T = Ø. We must show that either S = Y or T = Y. Since Y is connected, it suffices to show that S and T are each open.

Let y Y and let U be an open neighborhood of f(y) that is evenly covered. Let V and W be sheets of p⁻¹(U) provided by the definition of covering map, such that g(y) V and h(y) W. If g(y) = h(y), then V = W, whereas if g(y) ≠ h(y), then V and W are disjoint.

Since g and h are continuous at y, there is an open neighborhood N of y such that g(N) ⊂ V and h(N) ⊂ W. If y T, then V ∩ W = Ø, so that g(z) ≠ h(z) for all z N, and N ⊂ T. It follows that T is open. On the other hand, if y S, then V = W. Furthermore, for each z N, g(z) must be that unique point ν V such that p(ν) = f(z) and h(z) must be that unique point w W such that p(w) = f(z). Since V = W, we conclude that ν = w, so that g = h on N, N ⊂ S, and S is also open.

5.2 Theorem (Path Lifting Theorem): Let p : E → X be a covering map, let γ : [0,1] → X be a path, and let e₀ E satisfy p(e₀) = γ(0). Then there exists a unique path α : [0,1] → E such that α(0) = e₀ and p∘α = γ.

Proof: For each x X, choose an open neighborhood U(x) of x that is evenly covered by p. The open sets γ⁻¹(U(x)), x X, form an open cover of [0,1]. Since [0,1] is compact, we can find 0 = s₀ < s₁ < . . . < s_m = 1 and evenly covered open sets U₁, . . . , U_m such that γ⁻¹(U_j) includes [s_{j − 1}, s_j], 1 ≤ j ≤ m. In other words,

γ([s_{j − 1, sj}]) ⊂ U_j, 1 ≤ j ≤ m.

The lift is performed now in m steps, as follows.

Since p(e₀) = γ(0) U₁, there is an open neighborhood V₁, of e₀ that is mapped homeomorphically by p onto U₁. Define α on the interval [0, s₁] such that α(s) is the unique point of V₁ covering γ(s). In other words, set

α = (p|ν₁)^−1_∘γ on [0, s₁].

Then α(0) = e₀ and pºα = γ on [0, s₁].

Now perform the same procedure, with e₀ = α(0) replaced by α(s₁) and U₁, replaced by U₂, to extend α to the interval [s₁, s₂]. After m steps, we shall have lifted the entire path α.

The uniqueness of the lifted path α follows from Lemma 5.1.

Actually, one can lift a family of paths depending continuously on a parameter, so that the lifted paths also depend continuously on the parameter. A version of this principle, which will suffice for our purposes, is as follows.

5.3 Theorem: Let p : E → X be a covering map, let F : [0,1] × [0,1] → X be a map, and let e₀ E satisfy p(e₀) = F(0,0). Then there exists a unique lift G : [0,1] × [0,1] → E of F such that G(0,0) = e₀.

Proof: The uniqueness again follows from Lemma 5.1.

According to the Path Lifting Theorem, there is a unique path t → e_t, 0 ≤ t ≤ 1, in E such that p(e_t) = F(0,t), 0 ≤ t ≤ 1. By the same theorem, there exists for each t a unique path s → G(s,t), 0 ≤ s ≤ 1, such that G(0,t) = e_t and p(G(s,t)) = F(s,t), 0 ≤ s ≤ 1. This defines the lift G of F. We must show that G is continuous on [0,1] × [0,1].

Let γ : [0,1] → X be the path defined by

γ(s) = F(s, 0), 0 ≤ s ≤ 1.

Consider the construction of the lift α(s) = G(s, 0) given in the proof of Theorem 5.2 and retain the notation of that proof. Since γ⁻¹(U_j) is an open neighborhood of [s_j−1, s_j], F⁻¹(U_j) includes [s_j−1, s_j] × [0, ε] for some small ε > 0. Consequently there exists ε > 0 such that

F([s_j−1, s_j] × [0, ε]) ⊂ U_j, 1 ≤ j ≤ m.

Since e₀ V₁, we can assume also that the initial points e_t belong to V₁ for 0 ≤ t ≤ ε since they depend continuously on t. Now, the procedure for constructing the lift shows that

G = (p|ν₁)^−1_∘F on [0, s₁] × [0, ε].

In particular, G is continuous on [0, s₁] × [0, ε]. Proceeding in this fashion, we find that G is continuous on each rectangle [s_j–1, s_j] × [0, ε], 1 ≤ j ≤ m, so that G is continuous on [0,1] × [0,ε].

The same proof shows that for each t₀ (0,1], there exists ε > 0 such that G is continuous on the rectangle [0,1] × [t₀ − ε, t₀ + ε]. (Replace t₀ + ε by t₀ if t₀ = 1.) Consequently G is continuous on [0,1] × [0,1].

Now let (E,e) and (X,b) be pointed spaces and let p : (E,e) → (X,b) be a covering map, that is, p : E → X is a covering map satisfying p(e) = b. Let γ : [0,1] → X be a loop based at b. By the Path Lifting Theorem, there is a unique lift α : [0,1] → E of γ that satisfies α(0) = e. The lift α need not be a loop since it need not terminate at e. However, the terminal point α(1) of α satisfies p(α(1)) = γ(1) = b, so that α(1) lies in the fiber p⁻¹(b) over b.

Suppose now that γ₁ is another loop in X based at b such that γ₁ is homotopic to γ with endpoints fixed. Let {γ_t}_0≤t≤1 be the homotopy, so that γ₀ = γ. Applying Theorem 5.3 to F(s,t) = γ_t(s), we obtain a map G : [0,1] × [0,1] → E such that G(0,0) = e and p(G(s,t)) = γ_t(s), 0 ≤ s, t ≤ 1. Then the path α_t in E defined by α_t(s) = G(s,t), 0 ≤ t ≤ 1, is a lift of γ_t and α₀ = α. We claim that the α_t’s all start at e. To see this, observe that the map t → G(0,t) is the unique lift to E, starting at e, of the constant path at b. Hence the lift coincides with the constant path at e and e = G(0,t) = α_t(0), 0 ≤ t ≤ 1. Similarly, the map t → G(1,t) is the unique lift to E, starting at G(1,0) = α(1), of the constant path at b. Hence α(1) = G(1,t) = α_t(1) for 0 ≤ t ≤ 1 and all of the paths α_t terminate at α(1). In particular, the lift α₁ of γ₁ to E starts at e and terminates at α(1).

We conclude that the terminal point α(1) is the same for all loops in the same homotopy class of γ. This allows us to define a function

Φ : π₁(X,b) → p⁻¹(b),

so that Φ([γ]) is the terminal point of the lift of γ to E that starts at e.

As an application, consider the figure eight X and its covering space E given in the illustration. The loop αβ in X lifts to a path in E that begins at e and terminates

at y₁ On the other hand, the loop αβ in X lifts to a path in E that begins at e and terminates at y₂. Since y₁ ≠ y₂, αβ is not homotopic to βα and

[α][β] ≠ [β][α].

Thus we arrive at the striking discovery that the fundamental group of the figure eight is not commutative. By applying the idea used above to more elaborate covering spaces, it is possible to prove Van Kampen’s Theorem, that the fundamental group of the figure eight is the group called the free group on two generators (Exercise 6).

5.4 Theorem: Let p : (E,e) → (X,b) be a covering map and suppose that E is simply connected. Then Φ is a one-to-one correspondence of π₁(X,b) and the fiber p⁻¹(b).

Proof: Suppose that y p⁻¹(b). Let α be a path in E from e to y and set γ = pºα. Then α is the lift of γ to E that starts at e, so that Φ([γ]) = α(1) = γ. Hence the function Φ is onto.

Suppose that γ₀ and γ₁ are loops in X based at b such that Φ([γ₀]) = Φ([γ₁]). Let α₀ and α₁ be lifts of γ₀ and γ₁ respectively, that start at e. Then α₀ and α₁ have the same terminal point, so that the path α₀α₁⁻¹ in E is a loop based at e. Since E is simply connected, there is a homotopy F : [0,1] × [0,1] → E of α₀α₁⁻¹ to the point e. Then p∘F : [0,1] × [0,1] → X is a homotopy of the loop γ₀γ₁⁻¹ to the point b. Hence [γ₀][γ₁]⁻¹ = [b] and [y₀] = [γ₁]. It follows that Φ is one-to-one.

As a corollary of the proof, we obtain the following.

5.5 Corollary: Let p : (E,e) → (X,b) be a covering map and suppose that E is simply connected. For each point y p⁻¹(b), let α_y be a path in E from e to y and let γ_y = p∘α_y be a loop in X based at b. If γ is any loop in X based at b, then there is a unique loop γ_y such that γ is homotopic to γ_y with endpoints fixed.

As an example, we apply Theorem 5.4 to the covering map

p : Sⁿ → Pⁿ

discussed earlier in this section. Let e be the north pole of Sⁿ and let b = p(e). Then p⁻¹(b) consists of two points, the north and south poles of Sⁿ. If n ≥ 2, then Sⁿ is simply connected (Exercise 3.1). Consequently Theorem 5.4 applies and π₁(Pⁿ, b) has exactly two elements. One element is the identity, and the other element is the homotopy class of p∘α, where α is any path on Sⁿ from the north pole to the south pole. Since any group with two elements is isomorphic to ₂, we obtain

π₁(Pⁿ) ≅ ₂, n ≥ 2.

Now we return to the prototypical covering map p : (,0) → (S¹, 1) given by (5.1). Since p⁻¹(1) coincides with the subset of consisting of the integers, the elements of the fundamental group π₁(S¹,1) are in one-to-one correspondence with the integers. We wish to show that this correspondence is a group isomorphism. For this, we consider in detail the procedure by which a loop in S¹ determines an integer.

Let γ be a loop in S¹ based at 1. Then the lift of γ is a map

h : [0,1] →

that satisfies

The terminal point h(1) of h is then an integer, which is called the index of γ and denoted by ind(γ). Thus ind(γ) is the element of p⁻¹(0) associated with [γ] by Theorem 5.4.

5.6 Theorem: Two loops α and β in S¹ based at 1 are in the same homotopy class if and only if they have the same index. The correspondence

[α] → ind(α)

is an isomorphism of π₁(S¹, 1) and the integers .

Proof: The only item that remains to be proved is that the correspondence is a homomorphism. To verify this, it suffices to show that

ind(α₁α₂) = ind(α₁) + ind(α₂)

whenever α₁ and α₂ are both loops in S¹ based at 1.

Choose maps h₁, h₂ : [0,1] → so that h₁(0) = h₂(0) = 0 and

α_j(S) = e^2πih_j(s), 0 ≤ s ≤ 1; j = 1,2.

Define

Then h : [0,1] → is continuous, h(0) = 0, and

(α₁α₂)(s) = e^2πih(s), 0 ≤ s ≤ 1.

Consequently

ind(α₁α₂) = h(1) = h₁(1) + h₂(1) = ind(α₁) + ind(α₂).

In Chapter IV we shall generalize the notion of index to mappings of the n-sphere. In that context, the integer associated with a map will be called the degree of the map. Thus ind(γ) coincides with the degree deg(γ) of γ, to be defined in Chapter IV. It is a historical accident that the terminology used for curves on the plane differs from that used in higher-dimensional topology.

EXERCISES

1. For m an integer, let α_m be the loop in S¹ defined by

α_m(s) = e^2πims, 0 ≤ s ≤ 1.

Show that every loop in S¹ based at 1 is homotopic with endpoints fixed to precisely one of the loops α_m.

2. Suppose that p : (E,e) → (X,b) is a covering map and that E is simply connected. Suppose furthermore that E and X are groups with identities e and b, respectively, and that p is a homomorphism. Suppose finally that for each fixed y E, the group multiplication z → y _* z is a continuous function on E. Prove that the fiber p⁻¹(b) is a subgroup of E, and that

π₁(X,b) ≅ p⁻¹(b).

3. Show that the exponential map p : ⁿ → Tⁿ, defined earlier in this section, is a covering map of ⁿ onto the n-torus Tⁿ. Show that

π₁(Tⁿ) ≅ ⊕ . . . ⊕ (n summands).

For each n-tuple (m₁, . . . , m_n) ⊕ . . . ⊕ , define explicitly a loop in Tⁿ based at (1, . . . ,1) in the corresponding homotopy class.

4. Show that the map p : → \{0}, defined by

p(z) = e^z, z ,

is a covering map. What is π₁(\{0})?

5. Show that the restriction of the map p of Exercise 4 to the horizontal strip E = {x + iy : c < y < d} is a covering map of E over the open annulus {w : e^c < |w| < e^d}. What is the fundamental group of the open annulus?

What is the fundamental group of a closed annulus?

6. Let X be the figure-eight space, and let α and β be the loops in X indicated in the figure and discussion preceding Theorem 5.4. Show that every element of π₁(X) can be expressed uniquely as a finite product

[α]^m₁[β]^m₂[α]^m₃. . . ,

where m₁m₂, . . . are integers and m_j ≠ 0 for j ≥ 2 (m₁ = 0 is allowed). Note: This proves Van Kampen’s Theorem, that π₁(X) is the free group with generators [α] and [β]. For the uniqueness assertion, construct an appropriate covering space of X.

7. A topological space Y is locally path-connected if for each y ε Y and neighborhood U of y, there exists a neighborhood V of y such that every point of V can be joined to y by a path in U. Let (Y,c) be a pointed topological space such that Y is locally path-connected and simply connected, let p : (E,e) → (X,b) be a covering map.

(a) Show that every map f : (Y,c) → (X,b) can be uniquely lifted to a map g : (Y,c) → (E,b).

(b) Suppose in addition that X is locally path-connected and E is simply connected. Show that if f is a covering map, then g is a homeo-morphism of Y and E. (In other words, a simply connected covering space of a locally path-connected space is unique.)

8. Let p : E → X be a covering map. A map is a covering transformation if p∘f = f.

(a) Show that with the operation of composition, the covering transformations form a group.

(b) Show that if X is locally path-connected and E is simply connected, then the group of covering transformations is isomorphic to the fundamental group of X.

9. What are the covering transformations of S² over P²? of over S¹?

10. Let X be a path-connected topological space such that every x X has an open neighborhood that is simply connected. Fix b ε X and let E be the set of all pairs (x, [γ]), where x X and γ is a path in X from b to x. For each simply connected open subset U of X and each path γ in X starting at b and terminating at some point γ(1) U, define W(U, γ) to be the set of all pairs (x, [γ][α]) in E such that x U and α is a path in U from γ(1) to x. Prove the following.

(a) The sets W(U, γ) form a base for a topology for E.

(b) The natural projection (x, [γ]) → x of E onto X is a covering map.

Remark: By Exercise 7, the space E is essentially unique. It is called the universal covering space of X.

11. Let X be the quotient space obtained from the union of circles {x² + y² = 1, z = 1/n} in ³, for 1 ≤ n ≤ ∞, by identifying the set {(1,0,1/n) : 1 ≤ n ≤ ∞} to a point b. Show that if p : (E,e) → (X,b) is a covering map, then E is not simply connected. (Thus X has no universal covering space. Why does Exercise 10 not apply?)

6. SOME APPLICATIONS OF THE INDEX

Recall that Bⁿ is the closed unit ball in ⁿ. It will be convenient to identify ² and so that B² becomes the closed unit disc in the complex plane, with boundary circle S¹. The applications to be presented in this section are based on the following theorem.

6.1 Theorem: Let f be a map from B² to S¹ that satisfies f(1) = 1. Then the loop α, defined by

α(s) = f(e^2πis), 0 ≤ s ≤ 1,

has index zero:

ind(α) = 0.

Proof: Define a loop β in B² by

β(s) = e^2πis, 0 ≤, s ≤ 1.

Then α = f∘β. Since B² is convex, Corollary 4.2 shows that α is homotopic with endpoints fixed to the constant loop in S¹ at 1. By Theorem 5.6, ind(α) = 0.

The first application asserts that S¹ is not a retract of B².

6.2 Theorem: There is no map f of B² onto S¹ such that f(z) = z for all z S¹.

Proof: If there were such a map f, then the loop s → e^2πis, 0 ≤ s ≤ 1, in S¹ would have index zero, by Theorem 6.1. However, the index of the loop is 1.

There is another proof of Theorem 6.2, one which illustrates more clearly the method of algebraic topology (sometimes referred to as algebraic arrowology), that is, the converting of topological problems to problems in algebra.

Alternate Proof: Let be the inclusion map. By Theorem 4.3, the mappings in the following commutative diagram

generate the following commutative diagram of group homomorphisms:

Since f∘j is the identity map of S¹, (f∘j)_* = f_*∘j_* is the identity isomorphism of ≅ π₁(S¹, 1). However, since π(B², 1) = 0, both f_* and j_* are the zero homo-morphisms. Again we have reached a contradiction.

A point x X is a Fixed point of a map f : X → X if f(x) = x. In general, one does not expect a map to have a fixed point. For instance, the antipodal map z → − z of Sⁿ onto Sⁿ has no fixed points. In contrast, the celebrated Brouwer Fixed Point Theorem asserts that every map f : Bⁿ → Bⁿ has a fixed point. We prove this theorem in the special case where n = 2. The general case is treated in Chapter IV.

6.3 Theorem: Any map f : B² → B² has a fixed point.

Proof: Suppose that f has no fixed point. For each z B², let g(z) be the point of

S¹ at which the ray issuing from f(z) and passing through z leaves B². Then g is a continuous map from B² to S¹ (the proof is left as an exercise). Since g(z) = z if z S¹, we obtain a contradiction to Theorem 6.2.

The next application involves a preliminary reduction and then a more detailed analysis of the behavior of the index. Lurking in the background is projective space.

6.4 Theorem (Borsuk-Ulam Theorem): Let f be a map from S² to ². Then there exist antipodal points w and − w in S² such that f(w) = f(− w).

Proof: Define a map g : S² → ² by

g(w) = f(w) − f(− w), w S².

We must show that g vanishes at some point of S². The proof will depend only on the following property of g :

Consider the map h : B² → ² defined by

Here the nonnegative value of the square root is taken, so that h is obtained from g by flattening the top half of S² onto the closed disc B². From (6.1) we have

It suffices to show that any map h from B² to ² satisfying (6.2) vanishes at some point of B².

Suppose that such an h does not vanish on B². Then

defines a map φ : B² → S¹ which satisfies

By Theorem 6.1, the path

α(s) = φ(e^2πis), 0 ≤ s ≤ 1,

has index zero. We aim to obtain a contradiction by showing that the index of α is odd.

Choose k : [0,1] → such that

Then ind(α) = k(1). The condition (6.3) shows that

For each fixed s [0,1/2], the number

is then an integer. Since the function defined by (6.5) depends continuously on s and has discrete range, it is constant—equal to the integer m, say—so that

k(s + 1/2) – k(s) = m + 1/2, 0 ≤ s ≤ 1/2.

Then

an odd integer.

6.5 “Corollary”: At any given instant of time, there are two antipodal points on the surface of the earth at which the temperature and the wind speed are the same.

The next result concerns the division of volumes by planes. It derives its picturesque name from its interpretation as the assertion that it is possible, with a single knife stroke, to divide two pieces of bread and a piece of ham each into equal halves, no matter how irregular the three pieces or how askew their relative locations.

6.6 Theorem (Ham Sandwich Theorem): Let U, V, and W be three bounded connected open subsets of ³. Then there is a plane in ³. that divides each of the sets into two pieces of equal volume.