90 lines
2.8 KiB
Markdown
90 lines
2.8 KiB
Markdown
# 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 $g_*$
|