SET THEORY AND ELEMENTARY

BY DR. ATASI DEB RAY

DEPARTMENT OF MATHEMATICS WEST BENGAL STATE UNIVERSITY BERUNANPUKURIA, MALIKAPUR 24 PARAGANAS (NORTH) KOLKATA - 700126 WEST BENGAL, INDIA E-mail : [email protected] Chapter 6

Fundamental Group and its basic properties

Module 6

Applications

1 2

6.6 Applications

1 ∼ The result Π1(S ) = Z is helpful for proving certain well known theorems in two dimension. We discuss Brouwer’s fixed point theorem and Borsuk Ulam theorem (for dimension 2) to show how fundamental groups can be utilized to get an elegant proof for such nontrivial results. In what follows, D2 = {z ∈ C : |z| ≤ 1}. It is easy to see that S1 = {z ∈ C : |z| = 1} is the boundary of D2.

Definition 6.6.1. A function f : X → X is said to have a fixed point x ∈ X, if f(x) = x.

Theorem 6.6.1. (Brouwer’s fixed point theorem) Every continuous function f : D2 → D2 has a fixed point.

Proof. Suppose, f : D2 → D2 is a continuous function such that f(z) 6= z, for all z ∈ D2. Consider the ray (1 − t)f(z) + tz (t ≥ 0) that starts from f(z) and passes through z. This ray will meet the boundary S1 at some point, say g(z). Then we get a continuous function g : D2 → S1 given by z 7→ g(z) such that g(z) = z, for all z ∈ S1. Consider the following diagram, where i denotes the inclusion map.

g S1 →i D2 → S1

This sequence of continuous functions induces another sequence of homomorphisms between the respective fundamental groups :

∼ 1 i# 2 ∼ g# 1 ∼ Z = Π1(S ) → Π1(D ) = 0 → Π1(S ) = Z

1 1 By functorial properties of fundamental groups, g# ◦i# = (g◦i)# = (IS )# = IΠ1(S ), showing that identity map on Z is factoring through the trivial group 0, which is impossible. Hence such a g can not exist and so, there is some z ∈ D2 such that f(z) = z.

We now show how fundamental groups help to prove a very important and well known result, knnown as Borsuk Ulam Theorem. We prepare the basics now to utilize them for establishing the main result.

Definition 6.6.2. A point −x ∈ Sn is said to be an antipodal point (or an antipode) of x ∈ Sn. A map h : Sn → Sm is said to be an antipode preserving map if h(−x) = −h(x) for all x ∈ Sn.

1 1 Example 6.6.1. Let Rφ : S → S be a rotation by an angle φ (0 ≤ φ ≤ π). i.e., Rφ(cos θ, sin θ) = (cos (θ + φ), sin (θ + φ)). Then Rφ is antipode preserving. Because, for any point z = (cos θ, sin θ), −z = (cos (π + θ), sin (π + θ)) and so,

Rφ(cos (π + θ), sin (π + θ)) = (cos (π + θ + φ), sin (π + θ + φ)) = (− cos (θ + φ), − sin (θ + φ))

= −Rφ(cos θ, sin θ)

The following are very useful observations about antipode preserving maps: 3

• Composition of two antipode preserving maps is antipode preserving. • If h : S1 → S1 is homotopic to a constant map C : S1 → S1 by the H and 1 1 1 Rφ : S → S is a rotation, then Rφ ◦ h is homotopic to a constant map C by the homotopy Rφ ◦ H. In other words, h is null homotopic ⇒ Rφ ◦ h is null homotopic. Before we state the theorem, we define covering projection that we require to prove this result. We shall discuss in detail the concepts of covering spaces and covering projections in the following chapter. Definition 6.6.3. Let p : Xe → X be a continuous surjection. • An open set U of X is said to be evenly covered by p, if p−1(U) is expressed as disjoint union of open sets Vα of Xe, such that p restricted to each Vα is a homeomorphism of −1 each Vα onto U. The collection {Vα} is called the partition of p (U) into slices and each Vα is called a sheet over U. • If each x ∈ X has an open neighbourhood U in X such that U is evenly covered by p in Xe, then p is called the covering map (or, covering projection) and Xe is called the covering space of X. Such an open neighbourhood U of x is referred to as an admissible open set. Theorem 6.6.2. If h : S1 → S1 is a continuous and antipode preserving function then h is not nullhomotopic. 1 1 Proof. Suppose b0 = (1, 0) ∈ S and h(b0) = a0 ∈ S . If b0 = a0 then we proceed with the 1 1 function h. If not, then choose a rotation Rφ : S → S such that Rφ(h(b0)) = b0. Then 1 1 consider the map k = Rφ ◦h. In view of the previous observations, it follows that k : S → S is a continuous and antipode preserving map such that k(b0) = b0. For proving that h is not null homotopic, it is enough to show that k is not nullhomotopic. Consider a map q : S1 → S1 given by q(cos θ, sin θ) = (cos 2θ, sin 2θ). Then q is a closed, continuous surjection and hence a quotient map. The inverse image of any point w on S1 under the map q consists of the antipodal points z and −z of S1. i.e., w = q(z) = q(−z). Therefore, q(k(−z)) = q(−k(z)) = q(k(z)). So, by the property of quotient maps, q ◦ k induces a continuous map f : S1 → S1 such that f ◦ q = q ◦ k. S1 / S1 k q q   S1 / S1 f

It is to be noted that b0 = q(b0) = q(k(b0)) = f(q(b0)) = f(b0). Also, k(−b0) = −b0. Applying the fundamental group functor, we get the following commutative diagram:

∼ 1 1 ∼ Z = Π1(S , b0) / Π1(S , b0) = Z k#

q# q#   ∼ 1 1 ∼ Z = Π1(S , b0) / Π1(S , b0) = Z f# 4

Also, q is a covering projection: 1 1 2 1 2 S is covered by four open sets U1 = S ∩ {(x, y) ∈ R : y > 0}, U2 = S ∩ {(x, y) ∈ R : 1 2 1 2 −1 y < 0}, U3 = S ∩ {(x, y) ∈ R : x > 0} and U4 = S ∩ {(x, y) ∈ R : x < 0}. q (U1) is the disjoint union of the points on S1 belonging to the first and third quadrant of R2, each of which is homeomorphic to U1. Similarly, for U2, U3 and U4. 1 We observe that if α is any path from b0 to −b0 in S , then the loop β = q ◦ α represents a 1 nontrivial element of Π1(S , b0). Because, α is a lift of β which starts at b0 but does not end at b0. 1 Now, f#([β]) = [f ◦ β] = [f ◦ (q ◦ α)] = [q ◦ (k ◦ α)]. Since (k ◦ α) is a path in S that starts from b0 and ends at −b0,[q ◦ (k ◦ α)] is non-trivial. Hence, f# is a nontrivial homomorphism from an infinite cyclic group Z to itself. So, it is injective. Similarly, q# is injective so that f# ◦q# is injective. Using functorial property, q# ◦k# = f# ◦q# and therefore, k# is injective and hence nontrivial. Therefore, k is not null homotopic.

Theorem 6.6.3. There is no antipode preserving map g : S2 → S1.

Proof. If possible let there be a continuous antipode preserving function g : S2 → S1. Let 1 2 1 1 S be the equator of S . Then g|S1 : S → S is a continuous antipode preserving map. By 2 Theorem 6.6.2, g|S1 is not null homotopic. It is known that the upper hemisphere E of S is homeomorphic to the 2-disc D2 and D2 is contractible. Consider the continuous extension 1 g|E : E → S of g|S1 . Then ψ(homeo) g| D2 −→ E −→E S1 gives rise to the non-trivial induced homomorphism

2 (g|E )# 1 0 = Π1(D ) = Π1(E) −→ Π1(S )

which is impossible.

Theorem 6.6.4. (Borsuk Ulam Theorem) If f : S2 → R2 is a continuous function then there exists a point x of S2 such that f(−x) = f(x).

2 2 1 f(x)−f(−x) Proof. Suppose for all x ∈ S , f(x) 6= f(−x). Define g : S → S by g(x) = ||f(x)−f(−x)|| . Then g is a continuous function such that g(−x) = −g(x), for all x ∈ S2. By Theorem 6.6.3, it is not possible. Hence, there exists some x ∈ S2 such that f(x) = f(−x).

Remark 6.6.1. Brouwer’s Fixed point Theorem and Borsuk Ulam Theorem, both can be generalized for any n ∈ N. In case n = 1, purely topological arguments suffice. But in n higher dimensions, fundamental groups are inadequate, as we have seen that Π1(S ) = 0 for all n ≥ 2. So, for proving these results in higher dimensions, one requires the concept of groups instead.

Next is a non-tivial geometric phenomenone in R2, which follows from Borsuk Ulam Theorem (for dimension 2).

Theorem 6.6.5. (Bisection Theorem) If A1 and A2 are any two bounded polygonal regions on R2 then there exists a in R2 that bisects each of them. 5

Proof. Consider R2 as a subspace R2 × {1} of R3. To show that there exists a line L on 2 2 R × {1} that bisects both A1 and A2. Let p be a point on S . Consider a plane P in R3 passing through origin and having p as its unit normal. This plane divides R3 in two half-spaces. Let fi(p) denote the area of the portion of Ai that lies on the same side of P as does p, for i = 1, 2. If p is the unit vector k then fi(p) =area Ai and if p is the unit vector -k then fi(p) = 0. In any other case, the plane P intersects the plane R2 × {1} in a line L that splits it into two half planes and fi(p) is the area of that part of Ai that lies o one side of this line (for i = 1, 2). Replacing p by −p we get the same plane but the other half space. So, fi(p) is the area of the other portion of Ai that lies on the other side of P from p. Hence, fi(p) + fi(−p) = area of Ai. 2 2 Consider g : S → R given by g(p) = (f1(p), f2(p)). By Borsuk Ulam Theorem, there exists 2 some p0 ∈ S such that g(p0) = g(−p0). Therefore, fi(p0) = fi(−p0), for i = 1, 2. Then 1 fi(p0) = 2 area of Ai, i = 1, 2.