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content/Math4202/Math4202_L12.md

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+# Math4201 Topology II (Lecture 12)
+
+## Algebraic topology
+
+### 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.
+
+> [!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?
+
+#### 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 $x_0\in X$
+
+<details>
+<summary>Example of simply connected space</summary>
+
+Intervals are simply connected.
+
+---
+
+Any star-shaped is simply connected.
+
+---
+
+$S^1$ is not simply connected, but $n\geq 2$, then $S^n$ is simply connected.
+
+</details>
+
+#### Lemma for simply connected space
+
+In a simply connected space $X$, and two paths having the same initial and final points are path homotopic.
+
+<details>
+<summary>Proof</summary>
+
+Let $f,g$ be paths having the same initial and final points, then $f(0)=g(0)=x_0$ and $f(1)=g(1)=x_1$.
+
+Therefore $[f]*[\bar{g}]\simeq_p [e_{x_0}]$ (by simply connected space assumption).
+
+Then
+
+$$
+\begin{aligned}
+[f]*[\bar{g}]&\simeq_p [e_{x_0}]\\
+([f]*[\bar{g}])*[g]&\simeq_p [e_{x_0}]*[g]\\
+[f]*([\bar{g}]*[g])&\simeq_p [e_{x_0}]*[g]\\
+[f]*[e_{x_1}]&\simeq_p [e_{x_0}]*[g]\\
+[f]&\simeq_p [g]
+\end{aligned}
+$$
+
+</details>
+
+#### 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 $h_*([f])=[h\circ f]$.
+
+$h_*$ is called the group homomorphism induced by $h$ relative to $x_0$.
+
+<details>
+<summary>Check the homomorphism property</summary>
+
+$$
+\begin{aligned}
+h_*([f]*[g])&=h_*([f*g])\\
+&=[h_*[f*g]]\\
+&=[h_*[f]*h_*[g]]\\
+&=[h_*[f]]*[h_*[g]]\\
+&=h_*([f])*h_*([g])
+\end{aligned}
+$$
+
+</details>
+
+#### 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.
+
+<details>
+<summary>Proof</summary>
+
+Let $f$ be a loop based at $x_0$.
+
+$$
+\begin{aligned}
+k_*(h_*([f]))&=k_*([h\circ f])\\
+&=[k\circ h\circ f]\\
+&=[(k\circ h)\circ f]\\
+&=(k\circ h)_*([f])\\
+\end{aligned}
+$$
+
+</details>
+
+#### Corollary of composite of group homomorphism
+
+Let $\operatorname{id}:(X,x_0)\to (X,x_0)$ be the identity map. This induces $(\operatorname{id})_*:\Pi_1(X,x_0)\to \Pi_1(X,x_0)$.
+
+If $h$ is a homeomorphism with the inverse $k$, with
+
+$$
+k_*\circ h_*=(k\circ h)_*=(\operatorname{id})_*=I=(\operatorname{id})_*=(h\circ k)_*
+$$
+
+This induced $h_*: \Pi_1(X,x_0)\to \Pi_1(Y,y_0)$ is an isomorphism.
+
+#### Corollary for homotopy and group homomorphism
+
+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 $h_*=k_*$.
+
+$$
+h_*([f])=[h\circ f]\simeq_p[k\circ h]=k_*([f])
+$$