update

content/Math4201/Math4201_L18.md

@@ -108,8 +108,81 @@ We want to find $\hat{g}$ such that $g=\hat{g}\circ f$.
 
 If $x_1,x_2\in X$, such that $f(x_1)=f(x_2)$ and $g(x_1)\neq g(x_2)$, then we cannot find $\hat{g}$.
 
-#### Proposition
+#### Proposition for continuous and quotient maps
 
-Let $f$ and $g$ be as above. Moreover, for any $y\in Y$, all the points in $f^{-1}(y)$ are mapped to a single point by $g$. Then there is a unique continuous map $\hat{g}$ such that $g=\hat{g}\circ f$.
+Let $X,Y,Z$ be topological spaces. $p$ is a quotient map from $X$ to $Y$ and $g$ is a continuous map from $X$ to $Z$. 
 
-Continue next week.
+Moreover, if for any $y\in Y$, the map $g$ is constant on $p^{-1}(y)$, then there is a continuous map $f: Y\to Z$ satisfying $f\circ p=g$.
+
+<details>
+<summary>Proof</summary>
+
+For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
+
+Define $f(y)\coloneqq g(x)$.
+
+Note that this is well-defined and it doesn't depend on the specific choice of $x$ that $p(x)=y$ because $g$ is constant on $p^{-1}(y)$.
+
+Then we check that $f$ is continuous.
+
+Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
+
+Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
+
+Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
+
+Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
+
+</details>
+
+> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
+>
+> We may define $p^{-1}(y)$ as the equivalence class of $y$ if $p$ is defined using the equivalence relation. By definition $p^{-1}([x])$ is the element of $x$ that are $\sim x$.
+
+#### Additional to the proposition
+
+Note that $f$ is unique.
+
+It is not hard to see that $f$ is a quotient map if and only if $g$ is a quotient map. (check book for detailed proofs)
+
+#### Definition of saturated map
+
+Let $p:X\to Y$ be a quotient map. We say $A\subseteq X$ is **saturated** by $p$ if $A=p^{-1}(B)$ for some $B\subseteq Y$.
+
+Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
+
+#### Proposition for quotient maps from saturated sets
+
+Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
+
+Assume that $A$ is saturated by $p$.
+
+1. If $A$ is closed or open, then $q$ is a quotient map.
+2. If $p$ is closed or open, then $q$ is a quotient map.
+
+<details>
+<summary>Proof</summary>
+
+We prove 1 and assume that $A$ is open, (the closed case is similar).
+
+clearly, $q:A\to p(A)$ is surjective.
+
+In general, restricting the domain and the range of a continuous map is continuous.
+
+Since $A$ is saturated by $p$, then $p^{-1}(p(A))=A$ is open, so $p(A)$ is open because $p$ is a quotient map. Let $V\subseteq p(A)$ and $q^{-1}(V)\subseteq A$ is open. Then $q^{-1}(V)=p^{-1}(V)$.
+
+(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
+
+(ii) $p^{-1}(V)\subseteq q^{-1}(V)$: $x\in p^{-1}(V)\implies p(x)\in V\subseteq p(A)$. This implies that $x\in p^{-1}(p(A))=A$ since $A$ is saturated by $p$. Therefore $x\in q^{-1}(V)$.
+
+Since $A$ is open in $X$, any open subspace of $A$ is open in $X$. In particular, $q^{-1}(V)=p^{-1}(V)$ is open in $X$.
+
+Since $p$ is a quotient map, and $p^{-1}(V)$ is open in $X$, $V$ is open in $Y$. So $V\subseteq p(A)$ is open in $Y$.
+
+This shows $q$ is a quotient map.
+
+---
+
+We prove 2 next time...
+
+</details>

content/Math4201/Math4201_L19.md

@@ -0,0 +1,4 @@
+# Math4201 Topology I (Lecture 19)
+
+## Quotient topology
+

content/Math4201/_meta.js

@@ -21,5 +21,6 @@ export default {
     Math4201_L15: "Topology I (Lecture 15)",
     Math4201_L16: "Topology I (Lecture 16)",
     Math4201_L17: "Topology I (Lecture 17)",
-    Math4201_L18: "Topology I (Lecture 18)"
+    Math4201_L18: "Topology I (Lecture 18)",
+    Math4201_L19: "Topology I (Lecture 19)",
 }