380: For Quotient Map, Its Restriction on Open or Closed Saturated Domain and on Restricted Image Codomain Is Quotient Map

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A description/proof of that for quotient map, its restriction on open or closed saturated domain and on restricted image codomain is quotient map

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of quotient map.
• The reader knows a definition of saturated subset.
• The reader knows a definition of closed set.
• The reader knows a definition of subspace topology.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
• The reader admits the proposition that any open set on any open topological subspace is open on the base space.
• The reader admits the proposition that any continuous surjection between any topological spaces is a quotient map if any codomain subset is closed if its preimage is closed.
• The reader admits the proposition that any closed set on any closed topological subspace is closed on the base space.
• The reader admits the proposition that for any quotient topological space, any subset is closed if and only if the preimage of the subset under the quotient map is closed.

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Target Context

• The reader will have a description and a proof of the proposition that for any quotient map, its restriction on any open or closed saturated subset domain and on the restricted image codomain is a quotient map.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

For any quotient map, $f:T_1\to T_2$, and any open or closed saturated subset with respect to $f$, $S\subseteq T_1$, the restriction, $f{|}_S\to f(S)$, is a quotient map.

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2: Proof

$f{|}_S$ is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. $f{|}_S$ is surjective.

For any subset, $S^{\prime }\subseteq f(S)$, $(f{|}_S{)}^{-1}(S^{\prime })=f^{-1}(S^{\prime })\cap S$, because for any point, $p\in (f{|}_S{)}^{-1}(S^{\prime })$, $f{|}_S(p)=f(p)\in S^{\prime }$ and $p\in S$. For any point, $p\in f^{-1}(S^{\prime })\cap S$, $f(p)=f{|}_S(p)\in S^{\prime }$. $S^{\prime }\subseteq f(S)$, $f^{-1}(S^{\prime })\subseteq f^{-1}f(S)=S$, where the last equation is by the definition of saturated subset. So, $f^{-1}(S^{\prime })\cap S=f^{-1}(S^{\prime })$.

Let us suppose that $S$ is open. For any subset, $S^{\prime }\subseteq f(S)$, such that $(f{|}_S{)}^{-1}(S^{\prime })$ is open on $S$, is $S^{\prime }$ open on $f(S)$? $(f{|}_S{)}^{-1}(S^{\prime })=f^{-1}(S^{\prime })$ is open on $S$, and is open on $T_1$, by the proposition that any open set on any open topological subspace is open on the base space. So, $S^{\prime }$ is open on $T_2$, because $f$ is a quotient map. $S^{\prime }=S^{\prime }\cap f(S)$ is open on $f(S)$, by the definition of subspace topology.

Let us suppose that $S$ is closed. For any subset, $S^{\prime }\subseteq f(S)$, such that $(f{|}_S{)}^{-1}(S^{\prime })$ is closed on $S$, is $S^{\prime }$ closed on $f(S)$? $(f{|}_S{)}^{-1}(S^{\prime })=f^{-1}(S^{\prime })$ is closed on $S$, and is closed on $T_1$, by the proposition that any closed set on any closed topological subspace is closed on the base space. So, $S^{\prime }$ is closed on $T_2$, by the proposition that for any quotient topological space, any subset is closed if and only if the preimage of the subset under the quotient map is closed. $S^{\prime }=S^{\prime }\cap f(S)$ is closed on $f(S)$, by the definition of subspace topology. By the proposition that any continuous surjection between any topological spaces is a quotient map if any codomain subset is closed if its preimage is closed, $f{|}_S$ is a quotient map.

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References

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