fix alignment
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Trance-0
@@ -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
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#### Proposition for continuous and quotient maps
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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 $Z$.
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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 $g=\hat{g}\circ f$.
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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 $f\circ p=g$.
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<details>
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<summary>Proof</summary>
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For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
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Define $f(y)\coloneqq g(x)$.
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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 $p^{-1}(y)$.
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Then we check that $f$ is continuous.
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Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
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Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
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Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
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Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
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</details>
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> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
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>
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> 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 $\sim x$.
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#### Additional to the proposition
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Note that $f$ is unique.
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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)
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#### Definition of saturated map
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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 $B\subseteq Y$.
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Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
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#### Proposition for quotient maps from saturated sets
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Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
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Assume that $A$ is saturated by $p$.
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1. If $A$ is closed or open, then $q$ is a quotient map.
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2. If $p$ is closed or open, then $q$ is a quotient map.
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<details>
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<summary>Proof</summary>
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We prove 1 and assume that $A$ is open, (the closed case is similar).
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clearly, $q:A\to p(A)$ is surjective.
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In general, restricting the domain and the range of a continuous map is continuous.
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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 $q^{-1}(V)=p^{-1}(V)$.
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(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
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(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 $x\in q^{-1}(V)$.
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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 $X$.
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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 $Y$.
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This shows $q$ is a quotient map.
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---
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We prove 2 next time...
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</details>
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Continue next week.
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@@ -2,3 +2,83 @@
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## Quotient topology
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### More propositions
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#### Proposition for continuous and quotient maps
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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 $Z$.
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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 $f\circ p=g$.
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<details>
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<summary>Proof</summary>
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For any $y\in Y$, take $x\in X$ such that $p(x)=y$ (since $p$ is surjective).
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Define $f(y)\coloneqq g(x)$.
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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 $p^{-1}(y)$.
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Then we check that $f$ is continuous.
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Let $U\subseteq Z$ be open. Then we want to show that $f^{-1}(U)\subseteq Y$ is open.
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Since $p$ is a quotient map, this is equivalent to showing that $p^{-1}(f^{-1}(U))\subseteq X$ is open. Note that $p^{-1}(f^{-1}(U))=g^{-1}(U)$.
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Since $g$ is continuous, $g^{-1}(U)$ is open in $X$.
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Since $g^{-1}(U)$ is open in $X$, $p^{-1}(g^{-1}(U))$ is open in $Y$.
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</details>
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> In general, $p^{-1}(y)$ is called the **fiber** of $p$ over $y$. The $g$ must be constant on the fiber.
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>
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> 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 $\sim x$.
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#### Additional to the proposition
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Note that $f$ is unique.
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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)
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#### Definition of saturated map
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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 $B\subseteq Y$.
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Equivalently, if $x\in A$, then $p^{-1}(p(x))\subseteq A$.
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#### Proposition for quotient maps from saturated sets
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Let $p:X\to Y$ be a quotient map and $q$ be given by restriction of $p$ to $A\subseteq X$. $q:A\to p(A)$, $q(x)=p(x),x\in A$.
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Assume that $A$ is saturated by $p$.
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1. If $A$ is closed or open, then $q$ is a quotient map.
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2. If $p$ is closed or open, then $q$ is a quotient map.
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<details>
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<summary>Proof</summary>
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We prove 1 and assume that $A$ is open, (the closed case is similar).
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clearly, $q:A\to p(A)$ is surjective.
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In general, restricting the domain and the range of a continuous map is continuous.
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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 $q^{-1}(V)=p^{-1}(V)$.
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(i) $q^{-1}(V)\subseteq p^{-1}(V)$: $x\in q^{-1}(V)\implies q(x)\in V$. Then $p(x)=q(x)\in V$
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(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 $x\in q^{-1}(V)$.
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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 $X$.
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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 $Y$.
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This shows $q$ is a quotient map.
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---
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We prove 2 next time...
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</details>
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