Classnotes for September 13

From 0506Topology

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Study	Apr 17	Office Hours
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Classnotes for September 13

Table of contents

1 About the Course

2 The Brouwer Fixed-Point Theorem

3 Category Theory

4 Point Set Topology: Continuity

5 Scanned Notes

5.1 JPG, by Annat Koren
5.2 PDF, by Gabriel Lee

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About the Course

On an intuitive level, topology is associated with deformability, malleability (excluding from discussion, of course, the tearing and cutting of objects), whilst algebra is associated with rigidity. On this note one is often reminded of how, in topology, a coffee mug is considered equivalent to a doughnut. As we were reminded both in class and on the Course Syllabus, the purpose of this course is to explore the remarkable relationship between topology and algebra.

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The Brouwer Fixed-Point Theorem

To give a taste of what we will be able to do later on with the tools of algebraic topology, we took a look at a well-known purely topological result, the Brouwer fixed-point theorem for the 3-disc $D 3$ , whose proof requires just such tools. Remember that $D 3$ is the unit ball in $\mathbb{R}^3$ , i.e. $D^3 := \left \{ x \in \mathbb{R}^3 \colon \lVert x \rVert \le 1 \right \}$ .

Theorem (Brouwer fixed-point theorem for the 3-disc): Every continuous map $g \colon D^3 \to D^3$ has a fixed point, i.e. $\exists x \in D^3$ s.t. $g (x) = x$

The proof of this theorem can be divided into two lemmas. Before stating the first, we need the following definition:

Definition: Let $r \colon D^3 \to S^2$ , where $S^2 := \left \{ x \in \mathbb{R}^3 \colon \lVert x \rVert = 1 \right \} = \partial D^3$ . If $r|_{S^2} = id_{S^2}$ , then $r$ is said to be a retract. Equivalently, $r$ is a retract if the following diagram commutes:

N.B. Munkres (http://vig.pearsoned.ca/catalog/academic/product/0,4096,0131816292,00.html) and others would call what we call a retract a rectraction, and would call $S 2$ a retract instead.

Lemma 1: There is no continuous retract $r \colon D^3 \to S^2$ .

Intuitively, the lemma is seen to be true: a retract would have to tear the interior so as to pull everything towards the boundary, but a continuous function can neither tear nor cut. We postpone the rigorous proof for a bit, and turn our attention to the second lemma.

Lemma 2: Lemma 1 implies the Brouwer fixed-point theorem for the 3-disc.

Proof: It suffices to prove the contrapositive of the lemma. Assume that there exists some continuous $g \colon D^3 \to D^3$ such that $\forall x \in D^3 \ g(x) \ne x$ . Define $f \colon D^3 \to S^2$ as follows: $\forall x \in D^3$ , let $f (x) = g (x) + t (x - g (x))$ , where $t > 0$ is chosen such that $\lVert f(x) \rVert = 1$ . Assume for now that the choice of $t$ is unique. If $x \in S^2$ , it follows immediately that $t = 1$ , and hence $f (x) = x$ . Thus $f$ is a retract.

It remains to prove that $t$ is uniquely defined and that $f$ is continuous. Squaring both sides of the equation defining $t$ and rearranging, one finds that $t$ satisfies the following quadratic equation:

$\lVert x - g(x) \rVert^2 t^2 + 2 \left \langle x - g(x), g(x) \right \rangle t - \left ( 1 - \lVert g(x) \rVert^2 \right ) = 0$ .

Since $\lVert g(x) \rVert \le 1$ , the discriminant of the quadratic is non-negative, and the product of its roots is non-positive. The discriminant cannot be zero since $g(x) \ne x$ . Hence, the quadratic has a unique positive root $t$ , given by the expression

$t = \frac{-\left \langle x - g(x), g(x) \right \rangle + \sqrt{\left \langle x - g(x), g(x) \right \rangle^2 + \lVert x - g(x) \rVert^2 \left ( 1 - \lVert g(x)\rVert^2 \right )}}{\lVert x - g(x) \rVert^2}$

Since $g(x) \ne x$ , this expression for $t$ is always well defined, and indeed, $t$ is a continuous function of $x$ . Thus $f$ is a continuous retract, q.e.d.

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Category Theory

The only thing that remains to be proved in order to establish the Brouwer fixed-point theorem for the 3-disc is Lemma 1. In order to do that, however, we will need to borrow from the future, so to speak, and use the machinery of homology and the language of category theory. We first need some definitions - the following, whilst not absolutely rigorous, suffices for our present discussion:

Definition: A category is a collection of objects (e.g. sets, groups, vector spaces) and morphisms between them. These morphisms can be described as “structure-preserving maps” such that the following axioms hold:

1. Every object has an identity morphism;

2. Morphisms can be composed, i.e. for objects $a$ , $b$ , and $c$ , the morphisms $f : a \to b$ and $g: b \to c$ can be composed to form a morphism $g \circ f \colon a \to c$ ;

3. The composition of morphisms is associative, i.e. for morphisms $h : a \to b$ , $g \colon b \to c$ , and $f \colon c \to d$ , $f \circ \left (g \circ h \right ) = \left (f \circ g \right ) \circ h$ .

Definition: A functor from a category $C 1$ to a category $C 2$ is a map which sends objects in $C 1$ to objects in $C 2$ and morphisms in $C 1$ to morphisms in $C 2$ in such a way that preserves structure (i.e. compositions and identity morphisms are preserved).

It follows immediately from the definition that if one has a functor from a category $C 1$ to a category $C 2$ , then a diagram (i.e. diagram of objects and morphisms) in $C 1$ will be mapped to a corresponding diagram in $C 2$ . If the initial diagram commutes, then so too will its image under the functor. This fact is key in the proof of Lemma 1. The final piece of the puzzle is the following theorem from the future, which we will, of course, state without proof at this time:

Theorem: There exists a functor $H 2$ from the category of topological spaces and continuous functions to the category of abelian groups and group homomorphisms such that:

1. $H_2 \left(D^3 \right) \cong 0$ , the trivial group;

2. $H_2 \left(S^2 \right) \cong \mathbb{Z}$ .

With this result in hand, the proof of Lemma 1 is easy.

Proof of Lemma 1: Assume by contradiction that such a continuous retract $r$ exists. Then, in the category of topological spaces and continuous functions, we have this commutative diagram:

We apply the functor $H 2$ of the above theorem to obtain the following diagram in the category of abelian groups and group homomorphisms, likewise commutative:

Finally, applying the isomorphisms guaranteed by the same theorem, we obtain the following diagram, which is also commutative:

We see immediately that this last diagram cannot possibly be commutative, and so we have our desired contradiction, q.e.d.

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Point Set Topology: Continuity

We are all familiar with the notion of continuous functions of the real line.

Definition: A function $f \colon \mathbb{R}^m \to \mathbb{R}^n$ is said to be continuous if $\forall \epsilon > 0 \ \forall x \in \mathbb{R}^m \ \exists \delta > 0$ s.t. $\forall y \in \mathbb{R}^n \ \| x - y \| < \delta \Rightarrow \| f(x) - f(y) \| < \epsilon$ .

This definition of continuity requires a notion of distance to make sense. We wish to generalize this concept to functions between sets that are not necessarily subsets of $\mathbb{R} ^n$ . One way to achieve this is to generalize the above definition of continuity to apply to maps between metric spaces. The greatest level of generalization—the route which we shall pursue—is achieved by considering the notion of open sets.

We shall now generalize the standard defintion of an open subset of $\mathbb{R} ^n$

Definition: A set $U \subset \mathbb{R}^n$ is open if $\forall x \in U \ \exists \epsilon > 0$ s.t. $B_\epsilon(x) = \{y \in \mathbb{R}^n \colon |x - y| < \epsilon\} \subset U$ .

Intuitively, a set is open if for any point in the set, one can find a ball centered at that point which is wholly within the set. With this definition, we can find an alternative defintion for continuity.

Theorem: A function $f: \mathbb{R}^m \to \mathbb{R} ^n$ is continuous if and only if $\forall$ open $U \in \mathbb{R}^n \ f^{-1}(U) = \{x \in \mathbb{R}^m \colon f(x) \in U\}$ is also open.

Thus, a continuous function is one under which the preimage of an open subset of the codomain is an open subset of the domain. With the above theorem, we have an alternative definition of continuity that does not explicitly mention distance, but rather, open sets.

By adopting this definition of continuity, determining what are the basic properties of open sets, and creating topological spaces to be sets with a collection of sets which satisify our axioms for openness, we may thus extend the idea of continuous function to spaces that are more general than the real numbers. The proof of the theorem requires no tricks and proceeds in an somewhat obvious fashion.

Proof: Assume $f \colon \mathbb{R} ^m \to \mathbb{R} ^n$ is continuous. Let $U \in \mathbb{R} ^n$ be any open set. Let $x \in f^{-1}(U)$ be an arbitrary point of the preimage of $U$ . Note that $f(x) \in U$ . By the openness of $U$ , we can find an open ball around $f (x)$ contained in $U$ . That is, $\exists \epsilon > 0$ s.t. $B_\epsilon(f(x)) \subset U$ . By the continuity of $f$ , $\exists \delta > 0$ s.t. $f(B_\delta(x)) \subset B_\epsilon(f(x)) \subset U$ . So, $B_\delta(x) \subset f^{-1}(U)$ and we have found, for an arbitrary point $x \in f^{-1}(U)$ an open ball centered at $x$ and wholly contained in $f - 1 (U)$ . Thus, $f - 1 (U)$ is open.

Assume $f - 1 (U)$ is open for any open set $U \in \mathbb{R}^n$ . Let $x \in \mathbb{R}^m$ and $ε > 0$ . $B ε (f (x))$ is an open subset of $\mathbb{R}^n$ . Taking $U = B ε (f (x))$ , $f - 1 (B ε (f (x)))$ is open by assumption. By the definition of open set, $\exists \delta > 0$ s.t. $f(B_\delta(x)) \subset B_\epsilon(f(x))$ . Therefore, $f$ is continuous, q.e.d.