NoteNextra-origin/content/Math4202/Math4202_L20.md

Math4202 Topology II (Lecture 20)

Algebraic Topology

Retraction and fixed point

Lemma of retraction

Let h:S^1\to X be a continuous map. The following are equivalent:

• h is null-homotopic (h is homotopic to a constant map).
• h extends to a continuous map k:B_1(0)\to X. (k|_{\partial B_1(0)}=h)

For this course, we use closed disk B_1(0)=\{(x,y)|d((x,y),(0,0))\leq 1\}.

• h_* is the trivial group homomorphism of fundamental groups (Image of \pi_1(S^1,x_0)\to \pi_1(X,x_0) is trivial group, identity).

Proof

First we will show that (1) implies (2).

By the null homotopic condition, there exists H:S\times I\to X that H(s,1)=h(s),H(s,0)=x_0.

Define the equivalence relation for all the point in the set H(s,0). Then there exists \tilde{H}:S\times I/\sim \to X is a continuous map. (by quotient map q:S\times I\to S\times I/\sim and H is continuous.)

Note that the cone shape is homotopic equivalent to the disk using polar coordinates. k|_{\partial B_1(0)}=H|_{S^1\times\{1\}}=h.

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Then we will prove that (2) implies (3).

Let i: S^1\to B_1(0) be the inclusion map. Then k\circ i=h, k_*\circ i_*=h_*:\pi_1(S^1,1)\to \pi_1(X,h(1)).

Recall that k:B_1(0)\to X, k_*:\pi_1(B_1(0),0)\to \pi_1(X,k(0)) is trivial, since B_1(0) is contractible.

Therefore k_*\circ i_*=h_* is the trivial group homomorphism.

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Now we will show that (3) implies (1).

Consider the map from \alpha:[0,1]\to S^1 by \alpha:x\mapsto e^{2\pi ix}. [\alpha] is a generator of \pi_1(S^1,1). (The lifting of \alpha to \mathbb{R} is 1, which is a generator of \mathbb{Z}.)

h\circ \alpha:[0,1]\to X is a loop in X representing h_*([\alpha]).

As h_*([\alpha]) is trivial, h\circ \alpha is homotopic to a constant loop.

Therefore, there exist a homotopy H:I\times I\to X, where H(s,0)= constant map, H(s,1)=h\circ \alpha(s).

Take \tilde{H}:S\times I/\sim \to X by \tilde{H}(\exp(2\pi is),0)=H(s,0), \tilde{H}(\exp(2\pi is),t)=H(s,t). x\in [0,1].

(From the perspective of quotient map, we can see that \alpha is the quotient map from I\times I to I\times I/\sim=S\times I. Then \tilde{H} is the induced continuous map from S\times I to X.)

Corollary of punctured plane

i:S^1\to \mathbb{R}^2-\{0\} is not null homotopic.

Proof

Recall from last lecture, r:\mathbb{R}^2-\{0\}\to S^1 is the retraction x\mapsto \frac{x}{|x|}.

Therefore, we have i_*:\pi_1(S^1,1)\to \pi_1(\mathbb{R}^2-\{0\},r(0)): \mathbb{Z}\to \pi_1(\mathbb{R}^2-\{0\},r(0)) is injective.

Therefore i_* is non trivial.

Therefore i is not null homotopic.