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Zheyuan Wu
2026-03-20 16:59:05 -05:00
parent cd8705ad9e
commit 3f4479157b
9 changed files with 250 additions and 27 deletions

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@@ -163,9 +163,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
#### Definition for equivalent classes of paths
$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
$$
[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}

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@@ -95,5 +95,5 @@ $\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$.
The fundamental group of $X$ at $x$ is defined to be
$$
(\Pi_1(X,x),*)
(pi_1(X,x),*)
$$

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@@ -8,9 +8,9 @@ The $*$ operation has the following properties:
#### Properties for the path product operation
Let $[f],[g]\in \Pi_1(X)$, for $[f]\in \Pi_1(X)$, let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$.
Let $[f],[g]\in pi_1(X)$, for $[f]\in pi_1(X)$, let $s:pi_1(X)\to X, [f]\mapsto f(0)$ and $t:pi_1(X)\to X, [f]\mapsto f(1)$.
Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in \Pi_1(X)$.
Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in pi_1(X)$.
This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$.
@@ -51,33 +51,33 @@ Let $x_0\in X$. A path starting and ending at $x_0$ is called a loop based at $x
The fundamental group of $X$ at $x$ is defined to be
$$
(\Pi_1(X,x),*)
(pi_1(X,x),*)
$$
where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
where $*$ is the product operation, and $pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
<details>
<summary>Example of fundamental group</summary>
Consider $X=[0,1]$, with subspace topology from standard topology in $\mathbb{R}$.
$\Pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$.
$pi_1(X,0)=\{e\}$, (constant function at $0$) since we can build homotopy for all loops based at $0$ as follows $H(s,t)=(1-t)f(s)+t$.
And $\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
And $pi_1(X,1)=\{e\}$, (constant function at $1$.)
---
Let $X=\{1,2\}$ with discrete topology.
$\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
$pi_1(X,1)=\{e\}$, (constant function at $1$.)
$\Pi_1(X,2)=\{e\}$, (constant function at $2$.)
$pi_1(X,2)=\{e\}$, (constant function at $2$.)
---
Let $X=S^1$ be the circle.
$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
$pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
</details>
@@ -85,7 +85,7 @@ A natural question is, will the fundamental group depends on the base point $x$?
#### Definition for $\hat{\alpha}$
Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:\Pi_1(X,x_0)\to \Pi_1(X,x_1)$ as follows:
Let $\alpha$ be a path in $X$ from $x_0$ to $x_1$. $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$. Define $\hat{\alpha}:pi_1(X,x_0)\to pi_1(X,x_1)$ as follows:
$$
\hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha]
@@ -93,12 +93,12 @@ $$
#### $\hat{\alpha}$ is a group homomorphism
$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$
$\hat{\alpha}$ is a group homomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$
<details>
<summary>Proof</summary>
Let $f,g\in \Pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
Let $f,g\in pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
$$
\begin{aligned}
@@ -129,4 +129,4 @@ The other case is the same
#### Corollary of fundamental group
If $X$ is path-connected and $x_0,x_1\in X$, then $\Pi_1(X,x_0)$ is isomorphic to $\Pi_1(X,x_1)$.
If $X$ is path-connected and $x_0,x_1\in X$, then $pi_1(X,x_0)$ is isomorphic to $pi_1(X,x_1)$.

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@@ -4,18 +4,18 @@
### Fundamental group
Recall from last lecture, the $(\Pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(\Pi_1(X,x_0),*)$ is isomorphic to $(\Pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.
Recall from last lecture, the $(pi_1(X,x_0),*)$ is a group, and for any two points $x_0,x_1\in X$, the group $(pi_1(X,x_0),*)$ is isomorphic to $(pi_1(X,x_1),*)$ if $x_0,x_1$ is path connected.
> [!TIP]
>
> How does the $\hat{\alpha}$ (isomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?
> How does the $\hat{\alpha}$ (isomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$) depend on the choice of $\alpha$ (path) we choose?
#### Definition of simply connected
A space $X$ is simply connected if
- $X$ is [path-connected](https://notenextra.trance-0.com/Math4201/Math4201_L23/#definition-of-path-connected-space) ($\forall x_0,x_1\in X$, there exists a continuous function $\alpha:[0,1]\to X$ such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$)
- $\Pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
- $pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
<details>
<summary>Example of simply connected space</summary>
@@ -59,7 +59,7 @@ $$
#### Definition of group homomorphism induced by continuous map
Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
@@ -80,7 +80,7 @@ $$
#### Theorem composite of group homomorphism
If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:\Pi_1(X,x_0)\to \Pi_1(Z,z_0)$ where $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$, $k_*:\Pi_1(Y,y_0)\to \Pi_1(Z,z_0)$,is a group homomorphism.
If $h:(X,x_0)\to (Y,y_0)$ and $k:(Y,y_0)\to (Z,z_0)$ are continuous maps, then $k_* \circ h_*:pi_1(X,x_0)\to pi_1(Z,z_0)$ where $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$, $k_*:pi_1(Y,y_0)\to pi_1(Z,z_0)$,is a group homomorphism.
<details>
<summary>Proof</summary>
@@ -100,7 +100,7 @@ $$
#### Corollary of composite of group homomorphism
Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.
Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:pi_1(X,x_0)\to pi_1(X,x_0)$.
If $h$ is a homeomorphism with the inverse $k$, with
@@ -108,7 +108,7 @@ $$
k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
$$
This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.
This induced $h_*: pi_1(X,x_0)\to pi_1(Y,y_0)$ is an isomorphism.
#### Corollary for homotopy and group homomorphism

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@@ -0,0 +1,78 @@
# 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 $\mathbb{R}^2-\{0\}$
</details>
#### Theorem for Deformation Retract
If $A$ is a deformation retract of $X$, then $A$ and $X$ have the same fundamental group.

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@@ -7,7 +7,7 @@
Consider the space of paths up to homotopy equivalence.
$$
\operatorname{Path}/\simeq_p(X) =\Pi_1(X)
\operatorname{Path}/\simeq_p(X) =pi_1(X)
$$
We want to impose some group structure on $\operatorname{Path}/\simeq_p(X)$.
@@ -33,9 +33,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
#### Definition for equivalent classes of paths
$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
$$
[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}
@@ -141,5 +141,5 @@ Continue next time.
The fundamental group of $X$ at $x$ is defined to be
$$
(\Pi_1(X,x),*)
(pi_1(X,x),*)
$$

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@@ -1,3 +1,5 @@
import { MathJax } from "nextra/components";
export default {
index: "Course Description",
"---":{
@@ -27,4 +29,5 @@ export default {
Math4202_L21: "Topology II (Lecture 21)",
Math4202_L22: "Topology II (Lecture 22)",
Math4202_L23: "Topology II (Lecture 23)",
Math4202_L24: "Topology II (Lecture 24)",
}