188 lines
6.4 KiB
Markdown
188 lines
6.4 KiB
Markdown
# Math4201 Topology I (Lecture 18)
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## Quotient topology
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Let $(X,\mathcal{T})$ be a topological space and $X^*$ be a set, $q:X\to X^*$ is a surjective map. The quotient topology on $X^*$:
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$U\subseteq X^*$ is open $\iff q^{-1}(U)$ is open in $X$.
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Equivalently,
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$Z\subseteq X^*$ is closed $\iff q^{-1}(Z)$ is closed in $X$.
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### Open maps
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Let $(X,\mathcal{T})$ and $(Y,\mathcal{T}')$ be two topological spaces
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Let $f:X\to Y$ is a quotient map if and only if $f$ is surjective and
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$U\subseteq Y$ is open $\iff f^{-1}(U)$ is open
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or equivalently
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$Z\subseteq Y$ is closed $\iff f^{-1}(Z)$ is closed.
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#### Definition of open map
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Let $X\to Y$ be **continuous**. We say $f$ is open if for any $V\subseteq X$ be open, $f(V)$ is open in $Y$.
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Let $X\to Y$ be **continuous**. We say $f$ is closed if for any $V\subseteq X$ be closed, $f(V)$ is closed in $Y$.
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$$
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ff^{-1}(U)=U\text{ if }f \text{ is surjective}=U\cap f(X)
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$$
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<details>
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<summary>Examples of open maps</summary>
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Let $X,Y$ be topological spaces. Define the projection map $\pi_X:X\times Y\to X$, $\pi_X(x,y)=x$.
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This is a surjective continuous map $(Y\neq \phi)$
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This map is open. If $U\subseteq X$ is open and $V\subseteq Y$ is open, then $U\times V$ is open in $X\times Y$ and such open sets form a basis.
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$\pi_X(U\times V)=\begin{cases}
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U&\text{ if }V\neq \emptyset\\
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\emptyset &\text{ if }V=\emptyset
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\end{cases}$
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In particular, image of any such open set is open. Since any open $W\subseteq X\times Y$ is a union of such open sets.
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$W=\bigcup_{\alpha\in I}U_\alpha\times V\alpha$
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$\pi_X(W)=\pi_X(\bigcup_{\alpha\in I}U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}\pi_X(U_\alpha\times V_\alpha)=\bigcup_{\alpha\in I}U_\alpha$
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is open in $X$.
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However, $\pi_X$ is not necessarily a closed map.
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Let $X=Y=\mathbb{R}$ and $X\times Y=\mathbb{R}^2$
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$Z\subseteq \mathbb{R}^2=\{(x,y)\in\mathbb{R}^2|x\neq 0, y=\frac{1}{x}\}$ is a closed set in $\mathbb{R}^2$
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$\pi_X(Z)=\mathbb{R}\setminus \{0\}$ is not closed.
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---
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Let $X=[0,1]\cup [2,3]$, $Y=[0,2]$ with subspace topology on $\mathbb{R}$
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Let $f:X\to Y$ be defined as:
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$$
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f(x)=\begin{cases}
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x& \text{ if } x\in [0,1]\\
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x-1& \text{ if }x\in [2,3]
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\end{cases}
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$$
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$f$ is continuous and surjective, $f$ is closed $Z\subseteq [0,1]\cup [2,3]=Z_1\cup Z_2$, $Z_1\subseteq [0,1],Z_2\subseteq [2,3]$ is closed, $f(Z)=f(Z_1)\cup f(Z_2)$ is closed in $X$.
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But $f$ is not open. Take $U=[0,1]\subseteq X$, $f=[0,1]\subseteq [0,2]$ is not open because of the point $1$.
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> In general, and closed surjective map is a quotient map. In particular, this is an example of a closed surjective quotient map which is not open.
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</details>
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Let $f$ be a surjective open map. Then $f$ is a quotient map:
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$U\subseteq Y$ is open and $f$ is continuous, $\implies f^{-1}(U)\subseteq X$ is open
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$f^{-1}(U)\subseteq X$ is open and $f$ is surjective and open, $\implies f(f^{-1}(U))=U$ is open.
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#### Proposition of continuous and open maps
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If $f$ is a continuous bijection, then $f$ is open. if and only if $f^{-1}$ is continuous.
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<details>
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<summary>Proof</summary>
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To show $f^{-1}$ is continuous, we have to show for $U\subseteq X$ open. $(f^{-1})^{-1}(U)=f(U)\subseteq Y$ is open.
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This is the same thing as saying that $f$ is open.
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</details>
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Let $f$ be a quotient map $f: X \to Y$, and $g$ be a continuous map $g:X\to Z$.
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We want to find $\hat{g}$ such that $g=\hat{g}\circ f$.
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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 $\hat{g}$.
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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> |