updates

content/Math4202/Math4202_L1.md

@@ -14,4 +14,10 @@ Classifying two dimensional surfaces.
 
 ## Quotient spaces
 
-Let $X$ be a topological space and $f:X\to Y$ is a continuous, surjective map. WIth the property that $U\subset Y$ is open if and only if $f^{-1}(U)$ is open in $X$, we say $f$ is a quotient map and $Y$ is a quotient space.
+Let $X$ be a topological space and $f:X\to Y$ is a
+
+1. continuous
+2. surjective map.
+3. With the property that $U\subset Y$ is open if and only if $f^{-1}(U)$ is open in $X$.
+
+Then we say $f$ is a quotient map and $Y$ is a quotient space.

content/Math4202/Math4202_L2.md

@@ -0,0 +1,62 @@
+# Math4202 Topology II (Lecture 2)
+
+## Reviewing quotient map
+
+Recall from last lecture example (Example 4 form Munkers):
+
+A map of wrapping closed unit circle to $S^2$, where $f:\mathbb{R}^2\to S^2$ maps everything outside of circle to south pole $s$.
+
+To show it is a quotient space, we need to show that $f$:
+
+1. is continuous (every open set in $S^2$ has reverse image open in $\mathbb{R}^2$)
+2. surjective (trivial)
+3. with the property that $U\subset S^2$ is open if and only if $f^{-1}(U)$ is open in $\mathbb{R}^2$.
+
+- If $A\subseteq S^2$ is open, then $f^{-1}(A)$ is open in $\mathbb{R}^2$. (consider the basis, the set of circle in $\mathbb{R}^2$, they are mapped to closed sets in $S^2$)
+
+- If $f^{-1}(A)$ is open in $\mathbb{R}^2$, then $A$ is open in $S^2$.
+  - If $s\notin A$, then $f$ is a bijection, and $A$ is open in $S^2$.
+  - If $s\in A$, then $f^{-1}(A)$ is open and contains the complement of set $S=\{(x,y)|x^2+y^2\geq 1\}=f^{-1}(\{s\})$, therefore there exists $U=\bigcup_{x\in S} B_{\epsilon _x}(x)$ is open in $\mathbb{R}^2$, $U\subseteq f^{-1}(A)$, $f^{-1}(\{s\})\subseteq U$.
+  - Since $\partial f^{-1}(\{s\})$ is compact, we can even choose $U$ to be the set of the following form
+  - $\{(x,y)|x^2+y^2>1-\epsilon\}$ for some $1>\epsilon>0$.
+  - So $f(U)$ is an open set in $A$ and contains $s$.
+  - $s$ is an interior point of $A$.
+  - Other oint $y$ in $A$ follows the arguments in the first case.
+
+### Quotient space
+
+#### Definition of quotient topology induced by quotient map
+
+If $X$ is a topological space and $A$ is a set and if $p:X\to A$ is surjective, there exists exactly one topology $\mathcal{T}$ on $A$ relative to which $p$ is a quotient map.
+
+$$
+\mathcal{T} \coloneqq \{U|f^{-1}(U)\text{ is open in }X\}
+$$
+
+and $\mathcal{T}$ is called the quotient topology on $A$ induced by $p$.
+
+#### Definition of quotient topology induced by equivalence relation
+
+Let $X$ be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p:X\to X^*$ be the surjective map that sends each $x\in X$ to the unique $A\in X^*$ such that each point of $X$ to the subset containing the point. In the quotient topology induced by $p$, the space $X^*$ is called the associated quotient space.
+
+<details>
+<summary>Example of quotient topology induced by equivalence relation</summary>
+
+Consider $S^n$  and $x\sim -x$, then the induced quotient topology is $\mathbb{R}P^n$ (the set of lines in $\mathbb{R}^n$ passing through the origin).
+
+</details>
+
+#### Theorem about a quotient map and quotient topology
+
+Let $p:X\to Y$ be a quotient map; and $A$ be a subspace of $X$, that is **saturated** with respect to $p$: Let $q:A\to p(A)$ be the restriction of $p$ to $A$.
+
+1. If $A$ is either open or closed in $X$, then $q$ is a quotient map.
+2. If $p$ is either open or closed, then $q$ is a quotient map.
+
+> [!NOTE]
+>
+> Recall the definition of saturated set:
+>
+> $\forall y\in Y$, consider the set $f^{-1}(\{y\})\subset X$, if $f^{-1}(\{y\})\cap A\neq \emptyset$, then $f^{-1}(\{y\})\subseteq A$. _sounds like connectedness_
+>
+> That is equivalent to say that $A$ is a union of $f^{-1}(\{y\})$ for some $y\in Y$.

content/Math4202/_meta.js

@@ -4,4 +4,5 @@ export default {
         type: 'separator'
     },
     Math4202_L1: "Topology II (Lecture 1)",
+    Math4202_L2: "Topology II (Lecture 2)",
 }

content/Math4302/Math4302_L2.md

@@ -0,0 +1,176 @@
+# Math4302 Modern Algebra (Lecture 2)
+
+## Recall from last lecture
+
+### Binary operations
+
+A binary operation that is not associative but commutative:
+
+Consider $(\mathbb{Z},*)$ where $a*b=|a-b|$.
+
+This is trivially commutative.
+
+But $a=4,b=3,c=1$ gives $(a*b)*c=(4*3)*1=1*1=0$. and $a*(b*c)=4*(3*1)=4*2=2$.
+
+#### Definition for identity element
+
+An element $e\in X$ is called identity element if $a*e=e*a=a$ for all $a\in X$.
+
+### Group
+
+#### Definition of group
+
+A group is a set $G$ with a binary operation $*$ that satisfies the following axioms:
+
+1. Closure: $\forall a,b\in G, a* b\in G$ (automatically guaranteed by definition of binary operation).
+2. Associativity: $\forall a,b,c\in G, (a* b)* c=a* (b* c)$.
+3. Identity element: $\exists e\in G, \forall a\in G, a* e=e* a=a$.
+4. Inverses: $\forall a\in G, \exists a^{-1}\in G, a* a^{-1}=a^{-1}* a=e$.
+
+> [!NOTE]
+>
+> The inverse of $a$ is unique: If there is $b'\in G$ such that $b'*a=a*b'=e$, then $b=b'$.
+>
+> Proof:
+>
+> $b'=b'*e=b'*(a*b)=(b'*a)*b=e*b=b$.
+>
+> apply the definition of group.
+
+<details>
+<summary>Example of group</summary>
+
+$(\mathbb{Z},+)$ is a group.
+
+$(\mathbb{Q},+)$ is a group.
+
+$(\mathbb{R},+)$ is a group.
+
+with identity $0$ and all abelian groups.
+
+---
+
+$(\mathbb{Z},\cdot)$, $\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ are not groups ($0$ has no inverse).
+
+---
+
+We can fix this by removing $0$.
+
+$(\mathbb{Q}\setminus\{0\},\cdot)$, $(\mathbb{R}\setminus\{0\},\cdot)$ are groups.
+
+---
+
+$(\mathbb{Z}\setminus\{0\},\cdot)$ is not a group.
+
+$(\mathbb{Z}_+,+)$ is not a group.
+
+---
+
+Consider $S$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.
+
+$(S,+)$
+
+- Identity: $f(x)=0$
+- Associativity: $(f+g)(x)=f(x)+g(x)$
+- Inverse: $f(x)=-f(x)$
+
+This is a group.
+
+$(S,\circ)$
+
+- Identity: $f(x)=x$
+- Associativity: $(f\circ g)(x)=f(g(x))$
+- Inverse: not all have inverse...... (functions which are not bijective don't have inverses)
+
+This is not a group.
+
+---
+
+$\operatorname{GL}_(n,\mathbb{R})$: set of $n\times n$ invertible matrices over $\mathbb{R}$.
+
+$(\operatorname{SL}_(n,\mathbb{R}),\cdot)$ where $\cdot$ is matrix multiplication.
+
+- Identity: $I_n$
+- Associativity: $(A\cdot B)\cdot C=A\cdot (B\cdot C)$
+- Inverse: $(A^{-1})^{-1}=A$
+
+This is a group.
+
+**Matrix multiplication is not generally commutative**, therefore it's not abelian.
+
+</details>
+
+#### Definition of abelian group
+
+A group $(G,*)$ is called abelian if $a* b=b* a$ for all $a,b\in G$. ($*$ is commutative)
+
+#### Properties of group
+
+1. $(a*b)^{-1}=b^{-1}* a^{-1}$
+
+<details>
+<summary>Proof</summary>
+
+$(b^{-1}* a^{-1})*(a*b)=b^{-1}* a^{-1}*a*b=b^{-1}* e*b=b*b^{-1}=e$
+
+$(a*b)* (b^{-1}* a^{-1})=a* b*b^{-1}* a^{-1}=a* e*a^{-1}=a*a^{-1}=e$
+
+</details>
+
+2. Cancellation from right and left:
+
+$$
+a*b=a*c\implies b=c
+$$
+
+$$
+b*a=c*a\implies b=c
+$$
+
+<details>
+<summary>Proof</summary>
+
+$$
+\begin{aligned}
+    a*b&=a*c\\
+    a^{-1}*(a*b)&=a^{-1}*(a*c)\\
+    e*b&=e*c\\
+    b&=c
+\end{aligned}
+$$
+
+right cancellation are the same
+
+</details>
+
+> [!NOTE]
+>
+> This also implies that every row/column of the table representation of the binary operation is distinct.
+
+3. We can solve equations $a*x=b \text{ and } x*a=b
+$ uniquely.
+
+$x=a^{-1}* b$, similarly $x=b* a^{-1}$.
+
+### Finite groups
+
+Group with 1 element $\{e\}$.
+
+Group with 2 elements $\{e,a\}$.
+
+And
+
+|*|e|a|
+|-|-|-|
+|e|e|a|
+|a|a|e|
+
+Group with 3 elements $\{e,a,b\}$.
+
+And the possible ways to fill the table are:
+
+|*|e|a|b|
+|-|-|-|-|
+|e|e|a|b|
+|a|a|b|e|
+|b|b|e|a|

content/Math4302/_meta.js

@@ -4,4 +4,5 @@ export default {
         type: 'separator'
     },
     Math4302_L1: "Modern Algebra (Lecture 1)",
+    Math4302_L2: "Modern Algebra (Lecture 2)",
 }