331: Map Between Topological Spaces Is Continuous iff Preimage of Each Closed Subset of Codomain Is Closed

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A description/proof of that map between topological spaces is continuous iff preimage of each closed subset of codomain 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 continuous map.
• The reader knows a definition of closed set.
• The reader admits the proposition that the preimage of the codomain minus any codomain subset under 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 any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain 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, $T_1,T_2$, and any map, $f:T_1\to T_2$, $f$ is continuous if and only if for each close subset, $C\in T_2$, the preimage, $f^{-1}(C)$, is closed on $T_1$.

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2: Proof

Let us suppose that $f^{-1}(C)$ is closed on $T_1$. For any open subset, $U\subseteq T_2$, $C$ can be taken to be $C:=T_2\setminus U$, closed on $T_2$. $f^{-1}(U)=f^{-1}(T_2\setminus C)=T_1\setminus f^{-1}(C)$, by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset, and as $f^{-1}(C)$ is closed on $T_1$, $f^{-1}(U)$ is open on $T_1$.

Let us suppose that $f$ is continuous. $U:=T_2\setminus C$ is open on $T_2$. $f^{-1}(C)=f^{-1}(T_2\setminus U)=T_1\setminus f^{-1}(U)$, 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, which is closed on $T_1$.

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References

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