120 lines
3.3 KiB
Markdown
120 lines
3.3 KiB
Markdown
# Math4201 Topology II (Lecture 12)
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## Algebraic topology
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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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> [!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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#### 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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<details>
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<summary>Example of simply connected space</summary>
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Intervals are simply connected.
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---
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Any star-shaped is simply connected.
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---
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$S^1$ is not simply connected, but $n\geq 2$, then $S^n$ is simply connected.
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</details>
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#### Lemma for simply connected space
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In a simply connected space $X$, and two paths having the same initial and final points are path homotopic.
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<details>
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<summary>Proof</summary>
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Let $f,g$ be paths having the same initial and final points, then $f(0)=g(0)=x_0$ and $f(1)=g(1)=x_1$.
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Therefore $[f]*[\bar{g}]\simeq_p [e_{x_0}]$ (by simply connected space assumption).
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Then
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$$
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\begin{aligned}
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[f]*[\bar{g}]&\simeq_p [e_{x_0}]\\
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([f]*[\bar{g}])*[g]&\simeq_p [e_{x_0}]*[g]\\
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[f]*([\bar{g}]*[g])&\simeq_p [e_{x_0}]*[g]\\
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[f]*[e_{x_1}]&\simeq_p [e_{x_0}]*[g]\\
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[f]&\simeq_p [g]
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\end{aligned}
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$$
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</details>
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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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$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
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<details>
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<summary>Check the homomorphism property</summary>
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$$
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\begin{aligned}
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h_*([f]*[g])&=h_*([f*g])\\
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&=[h_*[f*g]]\\
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&=[h_*[f]*h_*[g]]\\
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&=[h_*[f]]*[h_*[g]]\\
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&=h_*([f])*h_*([g])
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\end{aligned}
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$$
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</details>
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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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<details>
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<summary>Proof</summary>
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Let $f$ be a loop based at $x_0$.
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$$
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\begin{aligned}
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k_*(h_*([f]))&=k_*([h\circ f])\\
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&=[k\circ h\circ f]\\
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&=[(k\circ h)\circ f]\\
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&=(k\circ h)_*([f])\\
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\end{aligned}
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$$
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</details>
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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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If $h$ is a homeomorphism with the inverse $k$, with
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$$
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k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
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$$
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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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#### Corollary for homotopy and group homomorphism
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If $h,k:(X,x_0)\to (Y,y_0)$ are homotopic maps form $X$ to $Y$ such that the homotopy $H_t(x_0)=y_0,\forall t\in I$, then $h_*=k_*$.
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$$
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h_*([f])=[h\circ f]\simeq_p[k\circ h]=k_*([f])
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$$
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