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@@ -163,9 +163,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
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#### Definition for equivalent classes of paths
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$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
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$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
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On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
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On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
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$$
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[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}
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@@ -95,5 +95,5 @@ $\bar{f}_t=\bar{f}(1-ts)$ $s\in[\frac{1}{2},1]$.
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The fundamental group of $X$ at $x$ is defined to be
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$$
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(\Pi_1(X,x),*)
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(pi_1(X,x),*)
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$$
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@@ -8,9 +8,9 @@ The $*$ operation has the following properties:
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#### Properties for the path product operation
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Let $[f],[g]\in \Pi_1(X)$, for $[f]\in \Pi_1(X)$, let $s:\Pi_1(X)\to X, [f]\mapsto f(0)$ and $t:\Pi_1(X)\to X, [f]\mapsto f(1)$.
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Let $[f],[g]\in pi_1(X)$, for $[f]\in pi_1(X)$, let $s:pi_1(X)\to X, [f]\mapsto f(0)$ and $t:pi_1(X)\to X, [f]\mapsto f(1)$.
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Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in \Pi_1(X)$.
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Note that $t([f])=s([g])$, $[f]*[g]=[f*g]\in pi_1(X)$.
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This also satisfies the associativity. $([f]*[g])*[h]=[f]*([g]*[h])$.
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@@ -51,33 +51,33 @@ Let $x_0\in X$. A path starting and ending at $x_0$ is called a loop based at $x
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The fundamental group of $X$ at $x$ is defined to be
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$$
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(\Pi_1(X,x),*)
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(pi_1(X,x),*)
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$$
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where $*$ is the product operation, and $\Pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
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where $*$ is the product operation, and $pi_1(X,x)$ is the set o homotopy classes of loops in $X$ based at $x$.
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<details>
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<summary>Example of fundamental group</summary>
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Consider $X=[0,1]$, with subspace topology from standard topology in $\mathbb{R}$.
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$\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$.
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$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$.
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And $\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
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And $pi_1(X,1)=\{e\}$, (constant function at $1$.)
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---
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Let $X=\{1,2\}$ with discrete topology.
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$\Pi_1(X,1)=\{e\}$, (constant function at $1$.)
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$pi_1(X,1)=\{e\}$, (constant function at $1$.)
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$\Pi_1(X,2)=\{e\}$, (constant function at $2$.)
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$pi_1(X,2)=\{e\}$, (constant function at $2$.)
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---
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Let $X=S^1$ be the circle.
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$\Pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
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$pi_1(X,1)=\mathbb{Z}$ (related to winding numbers, prove next week).
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</details>
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@@ -85,7 +85,7 @@ A natural question is, will the fundamental group depends on the base point $x$?
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#### Definition for $\hat{\alpha}$
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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:
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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:
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$$
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\hat{\alpha}(\beta)=[\bar{\alpha}]*[f]*[\alpha]
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@@ -93,12 +93,12 @@ $$
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#### $\hat{\alpha}$ is a group homomorphism
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$\hat{\alpha}$ is a group homomorphism between $(\Pi_1(X,x_0),*)$ and $(\Pi_1(X,x_1),*)$
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$\hat{\alpha}$ is a group homomorphism between $(pi_1(X,x_0),*)$ and $(pi_1(X,x_1),*)$
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<details>
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<summary>Proof</summary>
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Let $f,g\in \Pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
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Let $f,g\in pi_1(X,x_0)$, then $\hat{\alpha}(f*g)=\hat{\alpha}(f)\hat{\alpha}(g)$
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$$
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\begin{aligned}
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@@ -129,4 +129,4 @@ The other case is the same
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#### Corollary of fundamental group
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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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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 @@
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### Fundamental group
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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.
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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.
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> [!TIP]
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>
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> 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?
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> 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?
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#### Definition of simply connected
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A space $X$ is simply connected if
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- $X$ 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 $\alpha(1)=x_1$)
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- $\Pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
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- $pi_1(X,x_0)$ is the trivial group for some $x_0\in X$
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<details>
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<summary>Example of simply connected space</summary>
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@@ -59,7 +59,7 @@ $$
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#### Definition of group homomorphism induced by continuous map
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Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:\Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
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Let $h:(X,x_0)\to (Y,y_0)$ be a continuous map, define $h_*:pi_1(X,x_0)\to pi_1(Y,y_0)$ where $h(x_0)=y_0$. by $h_*([f])=[h\circ f]$.
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$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
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@@ -80,7 +80,7 @@ $$
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#### Theorem composite of group homomorphism
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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.
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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.
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<details>
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<summary>Proof</summary>
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@@ -100,7 +100,7 @@ $$
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#### Corollary of composite of group homomorphism
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Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.
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Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:pi_1(X,x_0)\to pi_1(X,x_0)$.
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If $h$ is a homeomorphism with the inverse $k$, with
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@@ -108,7 +108,7 @@ $$
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k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
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$$
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This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.
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This induced $h_*: pi_1(X,x_0)\to pi_1(Y,y_0)$ is an isomorphism.
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#### Corollary for homotopy and group homomorphism
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78
content/Math4202/Math4202_L24.md
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78
content/Math4202/Math4202_L24.md
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@@ -0,0 +1,78 @@
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# Math4202 Topology II (Lecture 24)
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## Algebraic Topology
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### Deformation Retracts and Homotopy Type
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Recall from last lecture, let $h,k:(X,x_0)\to (Y,y_0)$ be continuous maps. If there exists a homotopy of $h,y$ such that $H:X\times I\to Y$ that $H(x_0,t)=y_0$.
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Then $h_*=k_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$.
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We can prove this by showing that all the loop $f:I\to X$ based at $x_0$, $h_*([f])=k_*([f])$.
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That is $[h\circ f]=[k\circ f]$.
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This is a function $I\times I \to Y$ by $(s,t)\mapsto H(f(s),t)$.
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We need to show that this is a homotopy between $h\circ f$ and $k\circ f$.
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#### Theorem
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The Inclusion map $j:S^n\to \mathbb{R}^n-\{0\}$ induces on isomorphism of fundamental groups
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$$
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j_*:\pi_1(S^n)\to \pi_1(\mathbb{R}^n-\{0\})
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$$
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The function is injective.
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> Recall we showed that $S^1\to \mathbb{R}-\{0\}$ is injective by $x\mapsto \frac{x}{|x|}$.
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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.
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<details>
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<summary>Proof</summary>
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**Homotopy is well defined**.
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Consider $H:(\mathbb{R}^n-\{0\})\times I\to \mathbb{R}^n-\{0\}$.
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Given $(x,t)\mapsto tx+(1-t)\frac{x}{\|x\|}$.
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Note that $(t-\frac{1-t}{\|x\|})x=0\implies t=0\land t=1$.
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So this map is well defined.
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**Base point is fixed**.
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On point $(1,0)$ (or anything on the sphere), $H(x,0)=x$.
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</details>
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#### Definition of deformation retract
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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.
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Equivalently, there exists a homotopy $H:X\times I\to X$ such that:
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- $H(x,0)=x$ forall $x\in X$
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- $H(a,t)=a$ for all $a\in A$, $t\in I$
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- $H(x,1)\in A$ for all $x\in X$
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Equivalently,
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$r:H(x,1):X\to A$ is a retract.
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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.)
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<details>
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<summary>Example of deformation retract</summary>
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$S^1$ is a deformation retract of $\mathbb{R}^2-\{0\}$
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</details>
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#### Theorem for Deformation Retract
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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 @@
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Consider the space of paths up to homotopy equivalence.
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$$
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\operatorname{Path}/\simeq_p(X) =\Pi_1(X)
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\operatorname{Path}/\simeq_p(X) =pi_1(X)
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$$
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We want to impose some group structure on $\operatorname{Path}/\simeq_p(X)$.
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@@ -33,9 +33,9 @@ Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
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#### Definition for equivalent classes of paths
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$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
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$pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
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On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
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On $pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
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$$
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[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}
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@@ -141,5 +141,5 @@ Continue next time.
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The fundamental group of $X$ at $x$ is defined to be
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$$
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(\Pi_1(X,x),*)
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(pi_1(X,x),*)
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$$
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@@ -1,3 +1,5 @@
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import { MathJax } from "nextra/components";
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export default {
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index: "Course Description",
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"---":{
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@@ -27,4 +29,5 @@ export default {
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Math4202_L21: "Topology II (Lecture 21)",
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Math4202_L22: "Topology II (Lecture 22)",
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Math4202_L23: "Topology II (Lecture 23)",
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Math4202_L24: "Topology II (Lecture 24)",
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}
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