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content/Math4202/Math4202_L6.md

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+# Math4202 Topology II (Lecture 5)
+
+## Manifolds
+
+### Imbedding of Manifolds
+
+#### Definition for partition of unity
+
+Let $\{U_i\}_{i=1}^n$ be a finite open cover of topological space $X$. An indexed family of **continuous** function $\phi_i:X\to[0,1]$ for $i=1,...,n$ is said to be a **partition of unity** dominated by $\{U_i\}_{i=1}^n$ if
+
+1. $\operatorname{supp}(\phi_i)=\overline{\{x\in X: \phi_i(x)\neq 0\}}\subseteq U_i$ (the closure of points where $\phi_i(x)\neq 0$ is in $U_i$) for all $i=1,...,n$
+2. $\sum_{i=1}^n \phi_i(x)=1$ for all $x\in X$ (partition of function to $1$)
+
+#### Existence of finite partition of unity
+
+Let $\{U_i\}_{i=1}^n$ be a **finite open cover** of a **normal** space $X$ (Every pair of closed sets in $X$ can be separated by two open sets in $X$).
+
+Then there exists a partition of unity dominated by $\{U_i\}_{i=1}^n$.
+
+_A more generalized version, If the space is paracompact, then there exists a partition of unity dominated by $\{U_i\}_{i\in I}$ with locally finite. (Theorem 41.7)_
+
+We will prove for the finite partition of unity.
+
+<details>
+<summary>Proof for finite partition of unity</summary>
+
+Some intuitions:
+
+By definition for partition of unity, consider the sets $W_i,V_i$ defined as
+
+$$
+W_i=f^{-1}_i((\frac{1}{2n},1])\subseteq f^{-1}_i([\frac{1}{2n},1])\subseteq V_i=f^{-1}_i((0,1])\subseteq \operatorname{supp}(f_i)\subseteq U_i
+$$
+
+$$
+V_1\subseteq \overline{V_1}\subseteq U_1
+$$
+
+Note that $V_i$ is open and $\overline {V_i}\subseteq U_i$.
+
+And $\bigcup_{i=1}^n V_i=X$.
+
+and $W_i$ is open and $\overline{W_i}\subseteq V_i$.
+
+And $\bigcup_{i=1}^n W_i=X$.
+
+---
+
+**Step 1**: 
+
+$\exists$ V_i$ ope subsets $i=1,\dots,n$ such that $\overline{V_i}\subseteq U_i$, and $\bigcup_{i=1}^n V_i=X$.
+
+For $i=1$, consider $A_1=X-(U_2\cup U_3\cup \dots \cup U_n)$. Therefore $A_1$ is closed, and $A_1\cup U_1=X$.
+
+So $A_1\subseteq U_1$.
+
+Note that $A_1$ and $X-U_1$ are disjoint closed subsets of $X$.
+
+Since $X$ is normal, we can separate disjoint closed subsets $A_1$ and $X-U_1$.
+
+So we have $A_1\subset V_1\subseteq \overline{V_1}\subseteq U_1$ (by [normal space proposition](https://notenextra.trance-0.com/Math4201/Math4201_L37/#proposition-of-normal-spaces)).
+
+For $i=2$, note that $V_1\cup\left( \bigcup_{i=2}^n U_i\right)=X$,
+
+Take $A_2=X-\left(V_1\cup\left( \bigcup_{i=3}^n U_i\right)\right)$ (skipping $U_2$).
+
+Then we have $V_2\subseteq \overline{V_2}\subseteq U_2$.
+
+For $i=j$, we have
+
+$$
+A_j=X-\left(\left(\bigcup_{i=1}^{j-1}V_i\right)\cup \left(\bigcup_{i=j+1}^n U_i\right)\right)
+$$
+
+and $\bigcup_{i=1}^n V_i=X$.
+
+Repeat the above construction for $\{V_i\}_{i=1}^n$.
+
+Then we have $\{W_i\}_{i=1}^n$ open and $W_i\subseteq \overline{W_i}\subseteq V_i\subseteq \overline{V_i}\subseteq U_i$.
+
+And $\bigcup_{i=1}^n W_i=X$.
+
+**Step 2**:
+
+Using [Urysohn's lemma](https://notenextra.trance-0.com/Math4201/Math4201_L37/#urysohn-lemma). To construct the partition of unity $\phi_i$.
+
+> [!NOTE]
+>
+> Suppose
+> 
+> - $X$ be a normal space
+> - $Z_1,Z_2\subseteq X$ are closed
+> - $Z_1$ and $Z_2$ are disjoint
+>
+> Then:
+>
+> There exists $f:X\to[0,1]$ such that
+>
+> - $f(Z_1)=\{0\}$ and $f(Z_2)=\{1\}$
+> - $f$ is continuous.
+
+Since $W_1\subseteq \overline{W_1}\subseteq V_1\subseteq \overline{V_1}\subseteq U_1$,
+
+Note that $\overline{W_1}$ and $X-V_1$ are two disjoint closed subsets of normal space $X$
+
+Then we can have $f_1:X\to[0,1]$ such that $f_1(\overline{W_1})=\{0\}$ and $f_1(X-V_1)=\{1\}$.
+
+Then we have the remaining list of function $f_2,\dots,f_n$.
+
+Recall the definition for support of functions $\operatorname{supp}(f_i)=\overline{\{x\in X: f_i(x)>0\}}$. Since $f_i(x)=0$ for $x\in X-V_i$, we have $\operatorname{supp}(f_i)\subseteq \overline{V_i}$
+
+Next we need to check $\sum_{i=1}^n f_i(x)=1$ for all $x\in X$.
+
+Note that $\forall x\in X$, since $\bigcup_{i=1}^n W_i=X$, then there exists $i$ such that $x\in W_i$, thus $f_i(x)=1$.
+
+And $\sum _{i=1}^n f_i(x)\geq 1$.
+
+Then we do normalization for our value. Set $F(x)=\sum_{i=1}^n f_i(x)$.
+
+Since $F(x)$ is sum of continuous functions, $F$ is continuous.
+
+Then we define $\phi_i=f_i/F(x)$, since $F(x)\geq 1$, we are safe to divide by $F(x)$ and $\phi_i(x)$ is continuous.
+
+And $\operatorname{supp}(\phi_i)=\operatorname{supp}(f_i)\subseteq \overline{V_i}\subseteq U_i$.
+
+And $\sum_{i=1}^n \phi_i(x)=\frac{\sum_{i=1}^n f_i(x)}{F(x)}=\frac{F(x)}{F(x)}=1$ for all $x\in X$.
+
+</details>
+
+### Some Extension
+
+#### Definition of paracompact space
+
+Locally finite: $\forall x\in X$, $\exists$ open $x\in U$ such that $U$ only intersects finitely many open sets in $\mathcal{B}$.
+
+A space $X$ is paracompact if every open cover $A$ of $X$ has a **locally finite** refinement $\mathcal{B}$ of $A$ that covers $X$.
+
+## Algebraic Topology
+
+Homeomorphism: A topological space $X$ is homeomorphic to a topological space $Y$ if there exists a homeomorphism $f:X\to Y$
+
+- $f$ is continuous
+- $f^{-1}$ is continuous
+- $f$ is bijective
+
+Equivalence relation: If $\sim$ satisfies the following:
+
+- $\sim$ is reflexive $\forall x\in X, x\sim x$
+- $\sim$ is symmetric $\forall x,y\in X, x\sim y\implies y\sim x$
+- $\sim$ is transitive $\forall x,y,z\in X, x\sim y, y\sim z\implies x\sim z$
+
+Homeomorphism is an equivalence relation.
+
+- Reflexive: identity map
+- Symmetric: inverse map is also homeomorphism
+- Transitive: composition of homeomorphism is also homeomorphism
+
+Main Question: classify topological space up to homeomorphism.
+
+### Invariant in Mathematics
+
+Quantities associated with topological spaces that don't change under homeomorphism.
+
+We want to use some algebraic tools to classify topological spaces.

content/Math4202/_meta.js

@@ -8,4 +8,5 @@ export default {
     Math4202_L3: "Topology II (Lecture 3)",
     Math4202_L4: "Topology II (Lecture 4)",
     Math4202_L5: "Topology II (Lecture 5)",
+    Math4202_L6: "Topology II (Lecture 6)",
 }

content/Math4302/Math4302_L6.md

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+# Math4302 Modern Algebra (Lecture 6)
+
+## Subgroups
+
+### Dihedral group
+
+The dihedral group $D_n$ is the group of all rotations and reflections about the center of the regular polygon of $n$ sides.
+
+$|S_n|=n!, |D_n|=2n$
+
+### Cyclic group
+
+$G=\langle a\rangle=\{1,a,a^2,\cdots\}$ for some $a\in G$
+
+<details>
+<summary>Example of cyclic group</summary>
+
+$(\mathbb{Z}_n,+)$ is cyclic and $\mathbb{Z}_n=\langle 1\rangle=\{0,1,2,\cdots,n-1\}$
+
+---
+
+$(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle 1\rangle=\langle -1 \rangle$
+
+---
+
+$S_3$ is not cyclic
+
+$\langle e\rangle=\{e\}$
+$\langle (1,2)\rangle=\{e,(1,2)\}$
+$\langle (1,3)\rangle=\{e,(1,3)\}$
+$\langle (2,3)\rangle=\{e,(2,3)\}$
+$\langle (1,2,3)\rangle=\{e,(1,2,3),(1,3,2)\}$
+$\langle (1,3,2)\rangle=\{e,(1,3,2),(1,2,3)\}$
+
+</details>
+
+#### Every cyclic group is abelian
+
+Every cyclic group is abelian
+
+<details>
+<summary>Proof</summary>
+
+Let $G=\langle a\rangle$ be a cyclic group, then $\forall g_1,g_2\in G$ we have $g_1g_2=g_2g_1$ since $g_1g_2=a^k_1a^k_2=a^{k_1+k_2}$ and $g_2g_1=a^k_2a^k_1=a^{k_1+k_2}$
+
+</details>
+
+#### Definition for order of element
+
+Let $G$ be a group, then the order of $g\in G$ is defined to be the size of the smallest subgroup containing $g$.
+
+If $|\langle g\rangle|$ is infinite, then we say that $g$ has infinite order.
+
+<details>
+<summary>Example of order of element</summary>
+
+$5$ in $(\mathbb{Z},+)$ has infinite order.
+
+---
+
+$5$ in $(\mathbb{Z}_{10},+)$ has order $2$.
+
+$\langle 5\rangle=\{0,5\}$.
+
+---
+
+$5$ in $(\mathbb{Z}_{6},+)$ has order $6$.
+
+$\langle 5\rangle=\{0,5,4,3,2,1\}$.
+
+</details>
+
+#### Lemma for order of element
+
+Let $G$ be a group, then $a\in G$ has order $n$ if $n$ is the smallest positive integer such that $a^n=e$.
+
+<details>
+<summary>Proof</summary>
+
+There are 2 cases:
+
+Case 1:
+
+There is no positive $n$ such that $a^n=e$.
+
+Then $a^i\neq a^j$ if $i\neq j, i,j\in \mathbb{N}$.
+
+Reason: if $a^i=a^j$, then $a^{i-j}=e$.
+
+Then the order of group is infinite.
+
+Case 2:
+
+There is a positive $n$ such that $a^n=e$.
+
+Let $n$ be the smallest such positive integer. Then we claim $\langle a^n\rangle=\{e,a^1,a^2,\cdots,a^{n-1}\}$.
+
+We claim they are all distinct.
+
+Suppose not, then we can have $a^i=a^j$ for $i\neq j$, $0\leq i,j\leq n-1$.
+
+Then $a^{i-j}=e$ but $i-j\leq n-1$. Therefore $n$ is not the smallest positive integer such that $a^n=e$.
+
+</details>
+
+#### Theorem for cyclic group up to isomorphism
+
+Suppose $G$ is a cyclic group,
+
+- If $|G|=n$, then $|G|\simeq \mathbb{Z}_n^+$
+- If $|G|=\infty$, then $|G|\simeq \mathbb{Z}$.
+
+<details>
+<summary>Proof</summary>
+
+Case 1:
+
+If $|G|=\infty$, then we can map $G$ to $(\mathbb{Z},+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z},+)$.
+
+where $\phi(n+m)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.
+
+Case 2:
+
+If $|G|=n$, then we can map $G$ to $(\mathbb{Z}_n,+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z}_n,+)$.
+
+where $\phi(n+m)=\phi(r)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.
+</details>
+
+<details>
+<summary>Example</summary>
+
+Let $H=\langle (12)(345)\rangle\subseteq S_5$. Then $H\simeq \mathbb{Z}_6^+$.
+
+Let $\tau=(12)(345)$
+
+All the elements of $H$ are:
+
+- $\tau^0=(12)(345)$
+- $\tau^1=(453)$
+- $\tau^2=(12)(534)$
+- $\tau^3=(345)$
+- $\tau^4=(12)(453)$
+- $\tau^5=(534)$
+
+</details>
+
+#### GCD and order
+
+If $G=\langle a\rangle$, then $H=\langle a^k\rangle$, $|H|=\frac{n}{d}$ where $d=\operatorname{gcd}(n,k)$.

content/Math4302/_meta.js

@@ -8,4 +8,5 @@ export default {
     Math4302_L3: "Modern Algebra (Lecture 3)",
     Math4302_L4: "Modern Algebra (Lecture 4)",
     Math4302_L5: "Modern Algebra (Lecture 5)",
+    Math4302_L6: "Modern Algebra (Lecture 6)",
 }