updates

content/Math4202/Math4202_L23.md

@@ -94,3 +94,48 @@ For this section, we will show that $h_*$ is an isomorphism.
 #### Lemma for equality of homomorphism
 
 Let $h,k: (X,x_0)\to (Y,y_0)$ be continuous maps. If $h$ and $k$ are homotopic, and if **the image of $x_0$ under the homotopy remains $y_0$**. The homomorphism $h_*$ and $k_*$ from $\pi_1(X,x_0)$ to $\pi_1(Y,y_0)$ are equal.
+
+<details>
+<summary>Proof</summary>
+
+Let $H:X\times I\to Y$ be a homotopy from $h$ to $k$ such that
+$$
+H(x,0)=h(x), \qquad H(x,1)=k(x), \qquad H(x_0,t)=y_0 \text{ for all } t\in I.
+$$
+
+To show $h_*=k_*$, let $[f]\in \pi_1(X,x_0)$ be arbitrary, where
+$f:I\to X$ is a loop based at $x_0$, so $f(0)=f(1)=x_0$.
+
+Define
+$$
+F:I\times I\to Y,\qquad F(s,t)=H(f(s),t).
+$$
+Since $H$ and $f$ are continuous, $F$ is continuous. For each fixed $t\in I$, the map
+$$
+s\mapsto F(s,t)=H(f(s),t)
+$$
+is a loop based at $y_0$, because
+$$
+F(0,t)=H(f(0),t)=H(x_0,t)=y_0
+\quad\text{and}\quad
+F(1,t)=H(f(1),t)=H(x_0,t)=y_0.
+$$
+Thus $F$ is a based homotopy between the loops $h\circ f$ and $k\circ f$, since
+$$
+F(s,0)=H(f(s),0)=h(f(s))=(h\circ f)(s),
+$$
+and
+$$
+F(s,1)=H(f(s),1)=k(f(s))=(k\circ f)(s).
+$$
+
+Therefore $h\circ f$ and $k\circ f$ represent the same element of $\pi_1(Y,y_0)$, so
+$$
+[h\circ f]=[k\circ f].
+$$
+Hence
+$$
+h_*([f])=[h\circ f]=[k\circ f]=k_*([f]).
+$$
+Since $[f]$ was arbitrary, it follows that $h_*=k_*$.
+</details>