From ba9aedfc5a59abddb421bdde637cd5768fa218b3 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 28 Jan 2026 20:08:10 -0600
Subject: [PATCH] updates
---
content/Math4202/Math4202_L6.md | 2 +-
content/Math4202/Math4202_L7.md | 63 +++++++++++++++++++++++++
content/Math4202/_meta.js | 1 +
content/Math4302/Math4302_L7.md | 83 +++++++++++++++++++++++++++++++++
4 files changed, 148 insertions(+), 1 deletion(-)
create mode 100644 content/Math4202/Math4202_L7.md
create mode 100644 content/Math4302/Math4302_L7.md
diff --git a/content/Math4202/Math4202_L6.md b/content/Math4202/Math4202_L6.md
index e173b04..dc988eb 100644
--- a/content/Math4202/Math4202_L6.md
+++ b/content/Math4202/Math4202_L6.md
@@ -1,4 +1,4 @@
-# Math4202 Topology II (Lecture 5)
+# Math4202 Topology II (Lecture 6)
## Manifolds
diff --git a/content/Math4202/Math4202_L7.md b/content/Math4202/Math4202_L7.md
new file mode 100644
index 0000000..e8b32db
--- /dev/null
+++ b/content/Math4202/Math4202_L7.md
@@ -0,0 +1,63 @@
+# Math4202 Topology II (Lecture 6)
+
+## Algebraic Topology
+
+Classify 2-dimensional topological manifolds (connected) up to homeomorphism/homotopy equivalence.
+
+Use fundamental groups.
+
+We want to show that:
+
+1. The fundamental group is invariant under the equivalence relation.
+2. develop some methods to compute the groups.
+3. 2-dimensional topological spaces with the same fundamental group are equivalent (homeomorphism).
+
+### Homotopy of paths
+
+#### Definition of path
+
+If $f$ and $f'$ are two continuous maps from $X$ to $Y$, where $X$ and $Y$ are topological spaces. Then we say that $f$ is homotopic to $f'$ if there exists a continuous map $F:X\times [0,1]\to Y$ such that $F(x,0)=f(x)$ and $F(x,1)=f'(x)$ for all $x\in X$.
+
+The map $F$ is called a homotopy between $f$ and $f'$.
+
+We use $f\simeq f'$ to mean that $f$ is homotopic to $f'$.
+
+#### Definition of homotopic equivalence map
+
+Let $f:X\to Y$ and $g:Y\to X$ be two continuous maps. If $f\circ g:Y\to Y$ and $g\circ f:X\to X$ are homotopic to the identity maps $\operatorname{id}_Y$ and $\operatorname{id}_X$, then $f$ and $g$ are homotopic equivalence maps. And the two spaces $X$ and $Y$ are homotopy equivalent.
+
+> [!NOTE]
+>
+> This condition is weaker than homeomorphism. (In homeomorphism, let $g=f^{-1}$, we require $g\circ f=\operatorname{id}_X$ and $f\circ g=\operatorname{id}_Y$.)
+
+
+Example of homotopy equivalence maps
+
+Let $X=\{a\}$ and $Y=[0,1]$ with standard topology.
+
+Consider $f:X\to Y$ by $f(a)=0$ and $g:Y\to X$ by $g(y)=a$, where $y\in [0,1]$.
+
+$g\circ f=\operatorname{id}_X$ and $f\circ g=[0,1]\mapsto 0$.
+
+$g\circ f\simeq \operatorname{id}_X$
+
+and $f\circ g\simeq \operatorname{id}_Y$.
+
+Consider $F:X\times [0,1]\to Y$ by $F(a,0)=0$ and $F(a,t)=(1-t)y$. $F$ is continuous and homotopy between $f\circ g$ and $\operatorname{id}_Y$.
+
+This gives example of homotopy but not homeomorphism.
+
+
+
+#### Definition of null homology
+
+If $f:X\to Y$ is homotopy to a constant map. $f$ is called null homotopy.
+
+#### Definition of path homotopy
+
+Let $f,f':I\to X$ be a continuous maps from an interval $I=[0,1]$ to a topological space $X$.
+
+Two pathes $f$ and $f'$ are path homotopic if
+
+- there exists a continuous map $F:I\times [0,1]\to X$ such that $F(i,0)=f(i)$ and $F(i,1)=f'(i)$ for all $i\in I$.
+- $f(0)=f'(0)$ and $f(1)=f'(1)$.
\ No newline at end of file
diff --git a/content/Math4202/_meta.js b/content/Math4202/_meta.js
index 55468de..349f627 100644
--- a/content/Math4202/_meta.js
+++ b/content/Math4202/_meta.js
@@ -9,4 +9,5 @@ export default {
Math4202_L4: "Topology II (Lecture 4)",
Math4202_L5: "Topology II (Lecture 5)",
Math4202_L6: "Topology II (Lecture 6)",
+ Math4202_L7: "Topology II (Lecture 7)",
}
diff --git a/content/Math4302/Math4302_L7.md b/content/Math4302/Math4302_L7.md
new file mode 100644
index 0000000..1ceb228
--- /dev/null
+++ b/content/Math4302/Math4302_L7.md
@@ -0,0 +1,83 @@
+# Math4302 Modern Algebra (Lecture 7)
+
+## Subgroups
+
+### Cyclic group
+
+Last time, let $G$ be a group and $a\in G$. $|\langle a\rangle|=$ smallest positive $n$ such that $a^n=e$.
+
+$\langle a\rangle=\{a^0,a^1,a^2,\cdots,a^{n-1}\}$.
+
+
+#### Lemma subgroup of cyclic group is cyclic
+
+Every subgroup of a cyclic group is cyclic.
+
+$G=\langle a\rangle$.
+
+
+Proof
+
+Let $H\leq G$ be a subgroup.
+
+If $H=\{e\}$, we are done.
+
+Otherwise, let $m$ be the smallest positive integer such that $a^m\in H$. We claim $H=\langle a^m\rangle$.
+
+- $\langle a^m\rangle\subseteq H$. trivial since $a^m\in H$ and $H$ is a subgroup.
+- $H\subseteq\langle a^m\rangle$. Suppose $a^k\in H$, need to show $a^k\in \langle a^m\rangle$
+ Divide $k$ by $m$: $k=qm+r$, $0\leq r\leq m-1$, Then $a^k\in H\implies a^{qm+r}\in H$. Also $a^m\in H$, then $(a^m)^q\in H$, so $a^mq\in H$, $a^-mq\in H$, so $a^{k}a^{-mq}\in H$, so $a^r\in H$, so $r$ has to be zero.
+ By our choice of $m$, $k=mq$, so $a^k=a^mq\in \langle a^m\rangle$.
+
+
+
+
+Example
+
+Every subgroup of $(\mathbb{Z},+)$ is of the form
+
+like the multiples of $n$: $n\mathbb{Z}=\langle n\rangle$ for some $n\geq 0$.
+
+In particular, if $n,m\geq 1$ are in $\mathbb{Z}$, then the subgroup $\{nr+ms|r,s\in \mathbb{Z}\}\leq \mathbb{Z}$.
+
+is equal to $d\mathbb{Z}$ where $d=\operatorname{gcd}(n,m)$.
+
+
+
+Skip $\operatorname{gcd}$ part, check for Math 4111 notes in this site.
+
+
+#### Lemma for size of cyclic subgroup
+
+Let $G=\langle a\rangle$, $|G|=n$, and $H=\langle a^m\rangle\subseteq G$. Then $|H|=\frac{n}{d}$ where $d=\operatorname{gcd}(|G|,|H|)$.
+
+
+Proof
+
+Recall $|H|$ is the smallest power of $a^m$ which is equal to $e$.
+
+Let $d=\operatorname{gcd}(m,n)$, so $m=m_1d$, $n=n_1d$. and $\frac{n}{\operatorname{gcd}(m,n)}=n_1$,
+
+- $(a^m)^{n_1}=a^{mn_1}=a^{m_1dn_1}=a^{m_1n}=(a^n)^{m_1}=e$.
+- If $(a^m)^k=e$, the $a^{mk}=e\implies$ $mk$ is a multiple of $n$,
+ - If $a^\ell=e$, divide $\ell$ by $n$, $\ell=nq+r$, $0\leq r\leq n-1$, then $e=a^\ell=a^{nq+r}=a^r$, $r$ has to be zero, so $a^\ell=a^r=e$. $n|\ell$.
+- $n_1d|m_1dk$, but by the definition of smallest common divisor, $m_1,n_1$ should not have common divisor other than $1$. So $n_1|m_1k$, $n_1|k\implies k\geq n_1$.
+
+
+
+
+Example Applying the lemma
+
+Let $G=\langle a \rangle$, $|G|=6$, $H=\langle a^4\rangle$. Then $|H|=\frac{6}{d}=3$ where $d=\operatorname{gcd}(6,4)=2$.
+
+To check this we do enumeration $\langle a^4\rangle=\{e,a^4,a^2\}$.
+
+---
+
+Find generator of $\mathbb{Z}_9$:
+
+Using the coprime, we have $g=\{1,2,4,5,7,8\}$.
+
+
+
+Corollary: $\langle a^m\rangle=G\iff |H|=n\iff \frac{n}{d}=n\iff \operatorname{gcd}(m,n)=1$ $m,n$ are coprime.