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 </details>
 
-#### Lemma
+#### Lemma of homotopy equivalence
 
 Let $f,g:X\to Y$ be continuous maps. let $f(x_0)=y_0$ and $g(x_0)=y_1$. If $f$ and $g$ are homotopic, then there is a path $\alpha:I\to Y$ such that $\alpha(0)=y_0$ and $\alpha(1)=y_1$.
 
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+# Math4202 Topology II (Lecture 26)
+
+## Algebraic Topology
+
+### Deformation Retracts and Homotopy Type
+
+#### Lemma of homotopy equivalence
+
+Let $f,g:X\to Y$ be continuous maps. let
+
+$$
+f_*=\pi_1(X,f(x_0))\quad\text{and}\quad g_*=\pi_1(Y,g(x_0))
+$$
+
+And $H:X\times I\to Y$ is a homotopy from $f$ to $g$ with a path $H(x_0,t)=\alpha(t)$ for all $t\in I$.
+
+Then $\hat{\alpha}\circ f_*=[\bar{\alpha}*(f\circ \gamma)*\alpha]=[g\circ \gamma]=g_*$. where $\gamma$ is a loop in $X$ based at $x_0$.
+
+<details>
+<summary>Proof</summary>
+$I\times I\xrightarrow{\gamma_{id}} X\times I\xrightarrow{H} Y$
+
+- $I\times \{0\}\mapsto f\circ\gamma$
+- $I\times \{1\}\mapsto g\circ\gamma$
+- $\{0\}\times I\mapsto \alpha$
+- $\{1\}\times I\mapsto \alpha$
+
+As $I\times I$ is convex, $I\times \{0\}\simeq (\{0\}\times I)*(I\times \{1\})*(\{1\}\times I)$.
+
+</details>
+
+#### Corollary for homotopic continuous maps
+
+Let $h,k$ be homotopic continuous maps. And let $h(x_0)=y_0,k(x_0)=y_1$. If $h_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is injective, then $k_*:\pi_1(X,x_0)\to \pi_1(Y,y_1)$ is injective.
+
+<details>
+<summary>Proof</summary>
+
+$\hat{\alpha}$ is an isomorphism of $\pi_1(Y,y_0)$ to $\pi_1(Y,y_1)$.
+
+</details>
+
+#### Corollary for nulhomotopic maps
+
+Let $h:X\to Y$ be nulhomotopic. Then $h_*:\pi_1(X,x_0)\to \pi_1(Y,h(x_0))$ is a trivial group homomorphism (mapping to the constant map on $h(x_0)$).
+
+#### Theorem for fundamental group isomorphism by homotopy equivalence
+
+Let $f:X\to Y$ be a continuous map. Let $f(x_0)=y_0$. If $f$ is a [homotopy equivalence](https://notenextra.trance-0.com/Math4202/Math4202_L25/#definition-of-homotopy-equivalence) ($\exists g:Y\to X$ such that $fg\simeq id_X$, $gf\simeq id_Y$), then
+
+$$
+f_*:\pi_1(X,x_0)\to \pi_1(Y,y_0)
+$$
+is an isomorphism.
+
+<details>
+<summary>Proof</summary>
+
+Let $g:Y\to X$ be the homotopy inverse of $f$. 
+
+Then,
+
+$f_*\circ g_*=\alpha \circ id_{\pi_1(Y,y_0)}=\alpha$
+
+And $g_*\circ f_*=\bar{\alpha}\circ id_{\pi_1(X,x_0)}=\bar{\alpha}$
+
+So $f_*\circ (g_*\circ \hat{\alpha}^-1)=id_{\pi_1(X,x_0)}$
+
+And $g_*\circ (f_*\circ \hat{\alpha}^-1)=id_{\pi_1(Y,y_0)}$
+
+So $f_*$ is an isomorphism (have left and right inverse).
+</details>
+
+### Fundamental group of higher dimensional sphere
+
+$\pi_1(S^n,x_0)=\{e\}$ for $n\geq 2$.
+
+We can decompose the sphere to the union of two hemisphere and compute $\pi_1(S^n_+,x_0)=\pi_1(S^n_-,x_0)=\{e\}$
+
+But for $n\geq 2$, $S^n_+\cap S^n_-=S^{n-1}$, where $S^1_+\cap S^1_-$ is two disjoint points.
+
+#### Theorem for "gluing" fundamental group
+
+Suppose $X=U\cup V$, where $U$ and $V$ are open subsets of $X$. Suppose that $U\cap V$ is path connected, and $x\in U\cap V$. Let $i,j$ be the inclusion maps of $U$ and $V$ into $X$, the images of the induced homomorphisms
+
+$$
+i_*:\pi_1(U,x_0)\to \pi_1(X,x_0)\quad j_*:\pi_1(V,x_0)\to \pi_1(X,x_0)
+$$
+
+The image of the two map generate $\pi_1(X,x_0)$.