From b339bf50082fcc7f148fd8214b4b2e5cc4a3d765 Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 15 Oct 2025 14:20:00 -0500
Subject: [PATCH] fix alignment
---
content/Math4201/Math4201_L18.md | 77 +-----------------------------
content/Math4201/Math4201_L19.md | 80 ++++++++++++++++++++++++++++++++
2 files changed, 82 insertions(+), 75 deletions(-)
diff --git a/content/Math4201/Math4201_L18.md b/content/Math4201/Math4201_L18.md
index 4d839ec..1b71259 100644
--- a/content/Math4201/Math4201_L18.md
+++ b/content/Math4201/Math4201_L18.md
@@ -110,79 +110,6 @@ 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 ca
#### Proposition for continuous and quotient maps
-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 $Z$.
+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 $g=\hat{g}\circ f$.
-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 $f\circ p=g$.
-
-
-Proof
-
-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 $Y$.
-
-
-
-> 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.
-
-
-Proof
-
-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...
-
-
\ No newline at end of file
+Continue next week.
\ No newline at end of file
diff --git a/content/Math4201/Math4201_L19.md b/content/Math4201/Math4201_L19.md
index fe7e4ab..ae49ca6 100644
--- a/content/Math4201/Math4201_L19.md
+++ b/content/Math4201/Math4201_L19.md
@@ -2,3 +2,83 @@
## Quotient topology
+### More propositions
+
+#### Proposition for continuous and quotient maps
+
+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 $Z$.
+
+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 $f\circ p=g$.
+
+
+Proof
+
+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 $Y$.
+
+
+
+> 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.
+
+
+Proof
+
+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...
+
+
\ No newline at end of file