387: Map Between Topological Spaces Is Continuous if Domain Restriction of Map to Each Open Set of Open Cover is Continuous

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A description/proof of that map between topological spaces is continuous if domain restriction of map to each open set of open cover 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 any open set on any open topological subspace is open on the base space.

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Target Context

• The reader will have a description and a proof of the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover 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, $T_1,T_2$, and any map, $f:T_1\to T_2$, if there is an open cover of $T_1$, $\{U_{\alpha }\subseteq T_1\},{\cup }_{\alpha }{U}_{\alpha }=T_1$, such that each $f{|}_{U_{\alpha }}:U_{\alpha }\to T_2$ is continuous, $f$ is continuous.

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2: Proof

For any open set, $U\subseteq T_2$, ${f{|}_{U_{\alpha }}}^{-1}(U)$ is open on $U_{\alpha }$, and on $T_1$, by the proposition that any open set on any open subspace topological space is open on the base topological space. $f^{-1}(U)={\cup }_{\alpha }{f{|}_{U_{\alpha }}}^{-1}(U)$, because for any $p\in f^{-1}(U)$, $f(p)\in U$, but $p\in {\cup }_{\alpha }{U}_{\alpha }$, so, $f{|}_{U_{\alpha }}(p)\in U$ for an $\alpha$, $p\in {f{|}_{U_{\alpha }}}^{-1}(U)$; for any $p\in {\cup }_{\alpha }{f{|}_{U_{\alpha }}}^{-1}(U)$, $p\in {f{|}_{U_{\alpha }}}^{-1}(U)$ for an $\alpha$, so, $f{|}_{U_{\alpha }}(p)\in U$, so, $f(p)\in U$, so, $p\in f^{-1}(U)$. So, $f^{-1}(U)$ is open as the union of open sets.

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References

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