236: For Disjoint Union Topological Space, Inclusion from Constituent Topological Space to Disjoint Topological Space Is Continuous

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A description/proof of that for disjoint union topological space, inclusion from constituent topological space to disjoint topological space 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 disjoint union topology.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a description and a proof of the proposition that for any disjoint union topological space, the inclusion from any constituent topological space to the disjoint topological space 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 the set of any topological spaces, $\{T_{\alpha }\}$, and the disjoint union topological space, $T:=\coprod _{\alpha }{T}_{\alpha }$, each inclusion, $f_{\alpha }:T_{\alpha }\to T$, is continuous.

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2: Proof

For any open set, $U\subseteq T$, $U_{\alpha }:=U\cap T_{\alpha }$ is open on $T_{\alpha }$ by the definition of disjoint union topology. $(f_{\alpha }{)}^{-1}(U)=U_{\alpha }$, because for any $p\in (f_{\alpha }{)}^{-1}(U)$, $p\in T_{\alpha }$ and $p\in U$, so, $p\in U\cap T_{\alpha }$; for any $p\in U_{\alpha }$, $p\in T_{\alpha }$ and $p\in U$, $f_{\alpha }(p)\in U$, so, $p\in (f_{\alpha }{)}^{-1}(U)$. So, $(f_{\alpha }{)}^{-1}(U)$ is open on $T_{\alpha }$.

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References

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