# Math4202 Topology II (Lecture 25)

## Algebraic Topology

### Deformation Retracts and Homotopy Type

Recall from last lecture, Let $A\subseteq X$, if there exists a continuous map (deformation retraction) $H:X\times I\to X$ such that

- $H(x,0)=x$ for all $x\in X$
- $H(x,1)\in A$ for all $x\in X$
- $H(a,t)=a$ for all $a\in A$, $t\in I$

then the inclusion map$\pi_1(A,a)\to \pi_1(X,a)$ is an isomorphism.

<details>
<summary>Example for more deformation retract</summary>

Let $X=\mathbb{R}^3-\{0,(0,0,1)\}$.

Then the two sphere with one point intersect is a deformation retract of $X$.

---

Let $X$ be $\mathbb{R}^3-\{(t,0,0)\mid t\in \mathbb{R}\}$, then the cyclinder is a deformation retract of $X$.

</details>

#### Definition of homotopy equivalence

Let $f:X\to Y$ and $g:Y\to X$ be a continuous maps.

Suppose 

- the map $g\circ f:X\to X$ is homotopic to the identity map $\operatorname{id}_X$.
- the map $f\circ g:Y\to Y$ is homotopic to the identity map $\operatorname{id}_Y$.

Then $f$ and $g$ are **homotopy equivalences**, and each is said to be the **homotopy inverse** of the other.

$X$ and $Y$ are said to be **homotopy equivalent**.

<details>
<summary>Example</summary>

Consider the punctured torus $X=S^1\times S^1-\{(0,0)\}$.

Then we can do deformation retract of the glued square space to boundary of the square.

After glueing, we left with the figure 8 space.

Then $X$ is homotopy equivalent to the figure 8 space.

</details>

Recall the lemma, [Lemma for equality of homomorphism](https://notenextra.trance-0.com/Math4202/Math4202_L23/#lemma-for-equality-of-homomorphism)

Let $f:X\to Y$ and $g:X\to Y$, with homotopy $H:X\times I\to Y$, such that

- $H(x,0)=f(x)$ for all $x\in X$
- $H(x,1)=g(x)$ for all $x\in X$
- $H(x,t)=y_0$ for all $t\in I$, and $y_0\in Y$ is fixed.

Then $f_*=g_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is an isomorphism.

We wan to know if it is safe to remove the assumption that $y_0$ is fixed.

<details>
<summary>Idea of Proof</summary>

Let $k$ be any loop in $\pi_1(X,x_0)$.

We can correlate the two fundamental group $f\cric k$ by the function $\alpha:I\to Y$, and $\hat{\alpha}:\pi_1(Y,y_0)\to \pi_1(Y,y_1)$. (suppose $f(x_0)=y_0, g(x_0)=y_1$), it is sufficient to show that

$$
f\circ k\simeq \alpha *(g\circ k)*\bar{\alpha}
$$

</details>

#### Lemma

Let $f,g:X\to Y$ be continuous maps. let $f(x_0)=y_0$ and $g(x_0)=y_1$. If $f$ and $g$ are homotopic, then there is a path $\alpha:I\to Y$ such that $\alpha(0)=y_0$ and $\alpha(1)=y_1$.

Defined as the restriction of the homotopy to $\{x_0\}\times I$, satisfying $\hat{\alpha}\circ f_*=g_*$.

Imagine a triangle here:

- $\pi_1(X,x_0)\to \pi_1(Y,y_0)$ by $f_*$
- $\pi_1(Y,y_0)\to \pi_1(Y,y_1)$ by $\hat{\alpha}$
- $\pi_1(Y,y_1)\to \pi_1(X,x_0)$ by $g_*$