# Math4202 Topology II (Lecture 24)

## Algebraic Topology

### Deformation Retracts and Homotopy Type

Recall from last lecture, let $h,k:(X,x_0)\to (Y,y_0)$ be continuous maps. If there exists a homotopy of $h,y$ such that $H:X\times I\to Y$ that $H(x_0,t)=y_0$.

Then $h_*=k_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$.

We can prove this by showing that all the loop $f:I\to X$ based at $x_0$, $h_*([f])=k_*([f])$.

That is $[h\circ f]=[k\circ f]$.

This is a function $I\times I \to Y$ by $(s,t)\mapsto H(f(s),t)$.

We need to show that this is a homotopy between $h\circ f$ and $k\circ f$.

#### Theorem

The Inclusion map $j:S^n\to \mathbb{R}^n-\{0\}$ induces on isomorphism of fundamental groups

$$
j_*:\pi_1(S^n)\to \pi_1(\mathbb{R}^n-\{0\})
$$

The function is injective.

> Recall we showed that $S^1\to \mathbb{R}-\{0\}$ is injective by $x\mapsto \frac{x}{|x|}$.

We want to show that $j_*\circ r_*=id_{\pi_1(S^n)}\quad r_*\circ j_*=id_{\pi_1(\mathbb{R}^n-\{0\})}$, then $r_*$, $j_*$ are isomorphism.

<details>
<summary>Proof</summary>

**Homotopy is well defined**.

Consider $H:(\mathbb{R}^n-\{0\})\times I\to \mathbb{R}^n-\{0\}$.

Given $(x,t)\mapsto tx+(1-t)\frac{x}{\|x\|}$.

Note that $(t-\frac{1-t}{\|x\|})x=0\implies t=0\land t=1$.

So this map is well defined.

**Base point is fixed**.

On point $(1,0)$ (or anything on the sphere), $H(x,0)=x$.

</details>

#### Definition of deformation retract

Let $A$ be a subspace of $X$, we say that $A$ is a deformation retract of $X$ if the identity map of $X$ is homotopic to a map that carries all $X$ to $A$ such that each point of $A$ remains fixed during the homotopy.

Equivalently, there exists a homotopy $H:X\times I\to X$ such that:

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

Equivalently,

$r:H(x,1):X\to A$ is a retract.

If we let $j:A\to X$ be the inclusion map, then $r\circ j=id_A$, and $j\circ r\sim id_X$ (with $A$ fixed.)

<details>
<summary>Example of deformation retract</summary>

$S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$ 

---

Consider $\mathbb{R}^2-p=q$, the doubly punctured plane. "The figure 8" space is the deformation retract.

![Retraction of doubly punctured plane](https://notenextra.trance-0.com/Math4202/Retraction_of_doubly_punctured_plane.jpg)

</details>

#### Theorem for Deformation Retract

If $A$ is a deformation retract of $X$, then $A$ and $X$ have the same fundamental group.

