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)",
 }