Review section done

content/Math4201/Exam_reviews/Math4201_E1.md

@@ -306,3 +306,42 @@ A metric space $(Y,d)$ is bounded if there is $M\in \mathbb{R}^{\geq 0}$ such th
 
 Let $X$ be a topological space and $Y$ be a bounded metric space, then the set of all maps, denoted by $\operatorname{Map}(X,Y)$, $f:X\to Y\in \operatorname{Map}(X,Y)$ is a metric space with metric $\rho(f,g)=\sup_{x\in X} d(f(x),g(x))$.
 
+#### Space of continuous map is closed
+
+Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined above, then every continuous map is a limit point of some sequence of continuous maps.
+
+$$
+Z=\{f\in \operatorname{Map}(X,Y)|f\text{ is continuous}\}
+$$
+
+$Z$ is closed in $(\operatorname{Map}(X,Y),\rho)$.
+
+### Quotient space
+
+#### Quotient map
+
+Let $X$ be a topological space and $X^*$ is a set. $q:X\to X^*$ is a surjective map. Then $q$ is a quotient map.
+
+#### Quotient topology
+
+Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. Then
+
+$$
+\mathcal{T}^* \coloneqq \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
+$$
+
+is a topology on $X^*$ called quotient topology.
+
+$(X^*,\mathcal{T}^*)$ is called the quotient space of $X$ by $q$.
+
+#### Equivalent classes
+
+$\sim$ is a subset of $X\times X$ with the following properties:
+
+1. $x\sim x$ for all $x\in X$.
+2. If $(x,y)\in \sim$, then $(y,x)\in \sim$.
+3. If $(x,y)\in \sim$ and $(y,z)\in \sim$, then $(x,z)\in \sim$.
+
+The equivalence classes of $x\in X$ is denoted by $[x]=\{y\in X|y\sim x\}$.
+
+

content/Math4201/Math4201_L11.md

@@ -6,7 +6,7 @@
 >
 > A: No. Consider $X=[0,2\pi)$ and $Y=\mathbb{S}^1$ with standard topology in $\mathbb{R}^2$.
 >
-> Let $f\coloneq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ($\forall f^{-1}(V)$ is open in $X$)
+> Let $f\coloneqq \theta\in [0,2\pi)\to (\cos \theta, \sin \theta)\in \mathbb{S}^1$ is a continuous bijection. ($\forall f^{-1}(V)$ is open in $X$)
 >
 > But $f^{-1}$ is not continuous, consider the open set in $X, U=[0,\pi)$. Then $f^{-1}(U)=[0,\pi)$ is not open in $Y$.
 

content/Math4201/Math4201_L16.md

@@ -76,7 +76,7 @@ In fact, the metric topology by $d$ and $\overline{d}$ are the same. (proved in
 Let $X$ be a topological space. and $(Y,d)$ be a **bounded** metric space.
 
 $$
-\operatorname{Map}(X,Y)\coloneq \{f:X\to Y|f \text{ is a map}\}
+\operatorname{Map}(X,Y)\coloneqq \{f:X\to Y|f \text{ is a map}\}
 $$
 
 Define $\rho:\operatorname{Map}(X,Y)\times \operatorname{Map}(X,Y)\to \mathbb{R}$ by