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79 lines
2.2 KiB
Markdown
79 lines
2.2 KiB
Markdown
# 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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