87 lines
2.4 KiB
Markdown
87 lines
2.4 KiB
Markdown
# Math4201 Topology I (Lecture 17)
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## Quotient topology
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How can we define topologies on the space obtained points in a topological space?
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### Quotient map
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Let $(X,\mathcal{T})$ be a topological space. $X^*$ is a set and $q:X\to X^*$ is a surjective map.
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The quotient topology on $X^*$ is defined as follows:
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$$
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\mathcal{T}^* = \{U\subseteq X^*\mid q^{-1}(U)\in \mathcal{T}\}
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$$
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$U\subseteq X^*$ is open if and only if $q^{-1}(U)$ is open in $X$.
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In particular, $q$ is continuous map.
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#### Definition of quotient map
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$q:X\to X^*$ defined above is called a **quotient map**.
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#### Definition of quotient space
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$(X^*,\mathcal{T}^*)$ is called the **quotient space** of $X$ by $q$.
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### Typical way of constructing a surjective map
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#### Equivalence relation
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$\sim$ is a subset of $X\times X$ satisfying:
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- reflexive: $\forall x\in X, x\sim x$
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- symmetric: $\forall x,y\in X, x\sim y\implies y\sim x$
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- transitive: $\forall x,y,z\in X, x\sim y\text{ and } y\sim z\implies x\sim z$
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#### Equivalence classes
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Check equivalence relation.
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For $x\in X$, the equivalence class of $x$ is denoted as $[x]\coloneqq \{y\in X\mid y\sim x\}$.
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$X^*$ is the set of all equivalence classes on $X$.
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$q:X\to X^*$ is defined as $q(x)=[x]$ will be a surjective map.
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<details>
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<summary>Example of surjective maps and their quotient spaces</summary>
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Let $X=\mathbb{R}^2$ and $(s,t)\sim (s',t')$ if and only if $s-s'$ and $t-t'$ are both integers.
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This space as a topological space is homeomorphic to the torus.
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---
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Let $X=\{(s,t)\in \mathbb{R}^2\mid s^2+t^2\leq 1\}$ and $(s,t)\sim (s',t')$ if and only if $s^2+t^2$ and $s'^2+t'^2$. with subspace topology as a subspace of $\mathbb{R}^2$.
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This space as a topological space is homeomorphic to the spherical shell $S^2$.
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</details>
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We will show that the quotient topology is a topology on $X^*$.
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<details>
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<summary>Proof</summary>
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We need to show that the quotient topology is a topology on $X^*$.
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1. $\emptyset, X^*$ are open in $X^*$.
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$\emptyset, X^*$ are open in $X^*$ because $q^{-1}(\emptyset)=q^{-1}(X^*)=\emptyset$ and $q^{-1}(X^*)=X$ are open in $X$.
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2. $\mathcal{T}^*$ is closed with respect to arbitrary unions.
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$$
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q^{-1}(\bigcup_{\alpha \in I} U_\alpha)=\bigcup_{\alpha \in I} q^{-1}(U_\alpha)
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$$
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3. $\mathcal{T}^*$ is closed with respect to finite intersections.
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$$
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q^{-1}(\bigcap_{\alpha \in I} U_\alpha)=\bigcap_{\alpha \in I} q^{-1}(U_\alpha)
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$$
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</details>
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