205: Expansion of Continuous Map on Codomain Is Continuous

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A description/proof of that expansion of continuous map on codomain is continuous

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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 admits the proposition that in any nest of topological subspaces, the openness of any subset on any topological space does not depend on the superspace of which the topological space is regarded to be a subspace.

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Target Context

• The reader will have a description and a proof of the proposition that any expansion of any continuous map on the codomain is continuous.

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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, $T_1,T_2$, any topological subspaces, $T_3\subseteq T_4\subseteq T_2$, and any continuous map, $f:T_1\to T_3$, $f^{\prime }:T_1\to T_4$ as the expansion of $f$ on the codomain is continuous.

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2: Proof

For any open set, $U\subseteq T_4$, is ${f^{\prime }}^{-1}(U)$ open? ${f^{\prime }}^{-1}(U)={f^{\prime }}^{-1}(U\cap T_3)$, because $f^{\prime }(T_1)\subseteq T_3$. ${f^{\prime }}^{-1}(U\cap T_3)=f^{-1}(U\cap T_3)$, because $f^{\prime }$ and $f$ are practically the same. $U\cap T_3$ is open on $T_3$ by the definition of subspace topology no matter whether $T_3$ is regarded to be a subspace of $T_4$ or a subspace of $T_2$, by the proposition that in any nest of topological subspaces, the openness of any subset on any topological space does not depend on the superspace of which the topological space is regarded to be a subspace. As $f$ is continuous, $f^{-1}(U\cap T_3)$ is open, and so is ${f^{\prime }}^{-1}(U)$.

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References

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