From c57005e834a9b8e6a016589d2e854df0c2c847f8 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Thu, 10 Apr 2025 13:04:17 -0500
Subject: [PATCH] Update Math416_L23.md

---
 pages/Math416/Math416_L23.md | 80 ++++++++++++++++++++++++++++++++++++
 1 file changed, 80 insertions(+)

diff --git a/pages/Math416/Math416_L23.md b/pages/Math416/Math416_L23.md
index f5ed5e6..08ef4e1 100644
--- a/pages/Math416/Math416_L23.md
+++ b/pages/Math416/Math416_L23.md
@@ -86,4 +86,84 @@ QED
 
 ### Homotopy
 
+Suppose $\gamma_0, \gamma_1$ are two curves from
+$[0,1]$ to $\Omega$ with same end points $P,Q$.
+
+A homotopy is a continuous function of curves $\gamma_t, 0\leq t\leq 1$, deforming $\gamma_0$ to $\gamma_1$, keeping the end points fixed.
+
+Formally, if $H:[0,1]\times [0,1]\to \Omega$ is a continuous function satsifying
+
+1. $H(s,0)=\gamma_0(s)$, $\forall s\in [0,1]$
+2. $H(s,1)=\gamma_1(s)$, $\forall s\in [0,1]$
+3. $H(0,t)=P$, $\forall t\in [0,1]$
+4. $H(1,t)=Q$, $\forall t\in [0,1]$
+
+Then we say $H$ is a homotopy between $\gamma_0$ and $\gamma_1$. (If $\gamma_0$ and $\gamma_1$ are closed curves, $Q=P$)
+
+#### Lemma 9.12 Technical Lemma
+
+Let $\phi:[0,1]\times [0,1]\to \mathbb{C}\setminus \{0\}$ is continuous. Then there exists a continuous map $\psi:[0,1]\times [0,1]\to \mathbb{C}$ such that $e^\phi=\psi$. Moreover, $\psi$ is unique up to an additive constant in $2\pi i\mathbb{Z}$.
+
+Proof:
+
+Let $\phi_t(s)=\phi(s,t)$, $0\leq t\leq 1$.
+
+Then $\exists \psi_{00}$ such that $e^{\psi_{00}(s)}=\phi(0,t)$.
+
+$\exists \psi_{t}(s)$ such that $e^{\psi_{t}(s)}=\phi_t(s)$.
+
+We want to show $\psi_t(s)$ is continuous in $t$.
+
+Since $\exists \epsilon>0$ such that $\phi(s,t)$ is at least $\epsilon$ away from $0$ for all $s\in [0,1]$ and $t\in [0,1]$.
+
+Moreover, $\phi(s,t)$ is uniformly continuous.
+
+So $\exists \delta>0$ such that $|\phi(s,t)-\phi(s,t_0)|<\epsilon$ if $|t-t_0|<\delta$.
+
+Therefore,
+
+$$
+\begin{aligned}
+\left|\frac{\phi(s,t)}{\phi(s,t_0)}-1\right|&<\frac{\epsilon}{\phi(s,t_0)}
+&<1
+\end{aligned}
+$$
+
+So $\text {Re} \frac{\phi(s,t)}{\phi(s,t_0)}>0$.
+
+Therefore, $\text{Log} \frac{\phi(s,t)}{\phi(s,t_0)}=\chi(s,t)$ is continuous on $s\in [0,1], t\in [t_0-\delta, t_0+\delta]$.
+
+So $e^{\chi(s,t)}=\frac{\phi(s,t)}{\phi(s,t_0)}$, $\chi(s,t_0)=0,\forall s\in [0,1]$
+
+Define $\tilde{\psi}(s,t)=\chi(s,t)+\chi(s,t_0)$. So this function is continuous.
+
+And $e^{\tilde{\psi}(s,t)}=e^{\chi(s,t)+\chi(s,t_0)}=e^{\chi(s,t)}\cdot e^{\chi(s,t_0)}=\phi(s,t)$.
+
+$$
+\begin{aligned}
+\tilde{\psi}(0,t_0)&=\chi(0,t_0)+\psi(0,t_0) \\
+&=0+\psi_{00}(t_0) \\
+&=\psi_{00}(t_0)
+\end{aligned}
+$$
+
+$\tilde{\psi}(s,0)$ and $\psi(t,0)$ on $t\in[t_0-\delta, t_0+\delta]$ are both logs of the same function, and agree to each other on $t_0$.
+
+Therefore, $\tilde{\psi}(s,0)=\psi(s,0)+\text{const}$
+
+
+
+
+
+
+
+
+
+
+QED
+
+#### Theorem 9.13 Cauchy's Theorem for Homotopic Curves
+
+
+