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content/Math4202/Math4202_L9.md

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+# Math4202 Topology II (Lecture 9)
+
+## Algebraic Topology
+
+### Path homotopy
+
+Consider the space of paths up to homotopy equivalence.
+
+$$
+\operatorname{Path}/\simeq_p(X) =\Pi_1(X)
+$$
+
+We want to impose some group structure on $\operatorname{Path}/\simeq_p(X)$.
+
+Consider the $*$ operation on $\operatorname{Path}/\simeq_p(X)$.
+
+Let $f,g:[0,1]\to X$ be two paths, where $f(0)=a$, $f(1)=g(0)=b$ and $g(1)=c$.
+
+$$
+f*g:[0,1]\to X,\quad f*g(t)=\begin{cases}
+f(2t) & 0\leq t\leq \frac{1}{2}\\
+g(2t-1) & \frac{1}{2}\leq t\leq 1
+\end{cases}
+$$
+
+This connects our two paths.
+
+#### Definition for product of paths
+
+Given $f$ a path in $X$ from $x_0$ to $x_1$ and $g$ a path in $X$ from $x_1$ to $x_2$.
+
+Define the product $f*g$ of $f$ and $g$ to be the map $h:[0,1]\to X$.
+
+#### Definition for equivalent classes of paths
+
+$\Pi_1(X,x)$ is the equivalent classes of paths starting and ending at $x$.
+
+On $\Pi_1(X,x)$,, we define $\forall [f],[g],[f]*[g]=[f*g]$.
+
+$$
+[f]\coloneqq \{f_i:[0,1]\to X|f_0(0)=f(0),f_i(1)=f(1)\}
+$$
+
+#### Lemma
+
+If we have some path $k:X\to Y$ is a continuous map, and if $F$ is path homotopy between $f$ and $f'$ in $X$, then $k\circ F$ is path homotopy between $k\circ f$ and $k\circ f'$ in $Y$.
+
+If $k:X\to Y$ is a continuous map, and $f,g$ are two paths in $X$ with $f(1)=g(0)$, then
+
+$$
+(k\circ f)*(k\circ g)=k\circ(f*g)
+$$
+
+<details>
+<summary>Proof</summary>
+
+We check the definition of path homotopy.
+
+$k\circ F:I\times I\to Y$ is continuous.
+
+$k\circ F(s,0)=k(F(s,0))=k(f(s))=k\circ f(s)$.
+
+$k\circ F(s,1)=k(F(s,1))=k(f'(s))=k\circ f'(s)$.
+
+$k\circ F(0,t)=k(F(0,t))=k(f(0))=k(x_0$.
+
+$k\circ F(1,t)=k(F(1,t))=k(f'(1))=k(x_1)$.
+
+Therefore $k\circ F$ is path homotopy between $k\circ f$ and $k\circ f'$ in $Y$.
+
+---
+
+For the second part of the lemma, we proceed from the definition.
+
+$$
+(k\circ f)*(k\circ g)(t)=\begin{cases}
+k\circ f(2t) & 0\leq t\leq \frac{1}{2}\\
+k\circ g(2t-1) & \frac{1}{2}\leq t\leq 1
+\end{cases}
+$$
+
+and
+
+$$
+k\circ(f*g)=k(f*g(t))=k\left(\begin{cases}
+f(2t) & 0\leq t\leq \frac{1}{2}\\
+g(2t-1) & \frac{1}{2}\leq t\leq 1
+\end{cases}\right)=\begin{cases}
+k(f(2t))=k\circ f(2t) & 0\leq t\leq \frac{1}{2}\\
+k(g(2t-1))=k\circ g(2t-1) & \frac{1}{2}\leq t\leq 1
+\end{cases}
+$$
+
+</details>
+
+#### Theorem for properties of product of paths
+
+1. If $f\simeq_p f_1, g\simeq_p g_1$, then $f*g\simeq_p f_1*g_1$. (Product is well-defined)
+2. $([f]*[g])*[h]=[f]*([g]*[h])$. (Associativity)
+3. Let $e_{x_0}$ be the constant path from $x_0$ to $x_0$, $e_{x_1}$ be the constant path from $x_1$ to $x_1$. Suppose $f$ is a path from $x_0$ to $x_1$.
+    $$
+    [e_{x_0}]*[f]=[f],\quad [f]*[e_{x_1}]=[f]
+    $$
+    (Right and left identity)
+4. Given $f$ in $X$ a path from $x_0$ to $x_1$, we define $\bar{f}$ to be the path from $x_1$ to $x_0$ where $\bar{f}(t)=f(1-t)$.
+    $$
+    f*\bar{f}=e_{x_0},\quad \bar{f}*f=e_{x_1}
+    $$
+    $$
+    [f]*[\bar{f}]=[e_{x_0}],\quad [\bar{f}]*[f]=[e_{x_1}]
+    $$
+
+<details>
+<summary>Proof</summary>
+
+(1) If $f\simeq_p f_1$, $g\simeq_p g_1$, then $f*g\simeq_p f_1*g_1$.
+
+Let $F$ be homotopy between $f$ and $f_1$, $G$ be homotopy between $g$ and $g_1$.
+
+We can define
+
+$$
+F*G:[0,1]\times [0,1]\to X,\quad F*G(s,t)=\left(F(-,t)*G(-,t)\right)(s)=\begin{cases}
+F(2s,t) & 0\leq s\leq \frac{1}{2}\\
+G(2s-1,t) & \frac{1}{2}\leq s\leq 1
+\end{cases}
+$$
+
+$F*G$ is a homotopy between $f*g$ and $f_1*g_1$.
+
+We can check this by enumerating the cases from definition of homotopy.
+
+---
+
+Continue next time.
+
+</details>
+
+#### Definition for the fundamental group
+
+The fundamental group of $X$ at $x$ is defined to be
+
+$$
+(\Pi_1(X,x),*)
+$$