403: Subset of Quotient Topological Space Is Closed iff Preimage of Subset Under Quotient Map Is Closed

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A description/proof of that subset of quotient topological space is closed iff preimage of subset under quotient map 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 topology on set with respect to map.
• The reader knows a definition of quotient map.
• The reader knows a definition of closed set.
• 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 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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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 space, $T_1$, and any quotient topological space, $T_2$, with respect to any quotient map, $f:T_1\to T_2$, any subset, $C\subseteq T_2$, is closed if and only if $f^{-1}(C)$ is closed.

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2: Proof

Suppose that $f^{-1}(C)$ is closed. $f^{-1}(T_2\setminus C)=T_1\setminus f^{-1}(C)$ 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. $T_1\setminus f^{-1}(C)$ is open, so, $T_2\setminus C$ is open by the definition of quotient map, so, $C$ is closed.

Suppose that $C$ is closed. $T_2\setminus C$ is open. $f^{-1}(T_2\setminus C)=T_1\setminus f^{-1}(C)$ as before, and is open as the preimage of an open set under a continuous map, so, $f^{-1}(C)$ is closed.

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References

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