245: For Quotient Map, Codomain Subset Is Closed if Preimage of Subset Is Closed

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A description/proof of that for quotient map, codomain subset is closed if preimage of subset is closed

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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 topological space.
• The reader knows a definition of quotient map.
• The reader admits the proposition that the preimage of the codomain minus any codomain subset of any map is the domain minus the preimage of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any quotient map, any codomain subset is closed if the preimage of the subset is closed.

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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 topological spaces, $T_1$ and $T_2$, any quotient map, $f:T_1\to T_2$, and any subset, $S\subseteq T_2$, $S$ is closed if $f^{-1}(S)$ is closed.

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2: Proof

Suppose that $f^{-1}(S)$ is closed. $f^{-1}(T_2\setminus S)=T_1\setminus f^{-1}(S)$, by the proposition that the preimage of the codomain minus any codomain subset of any map is the domain minus the preimage of the subset. As $T_1\setminus f^{-1}(S)$ is open, by the definition of quotient map, $T_2\setminus S$ is open. So, $S$ is closed.

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References

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