Math4202 Topology II (Lecture 20)

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 $X$.)

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.