378: Characteristic Property of Disjoint Union

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A description/proof of characteristic property of disjoint union

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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, any map from the disjoint union topological space to any topological space is continuous if and only if the composition of the map after the inclusion from each constituent space to the disjoint union topological space is continuous, and the disjoint union topology is the unique topology that makes the disjoint union topological space have that property.

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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 disjoint union topological space, $T_1=\coprod _{\alpha \in A}{T}_{\alpha }$ where $\alpha \in A$ is any possibly uncountable indices set, any map to any topological space, $f:T_1\to T_2$, is continuous if and only if $f\circ {\iota }_{\alpha }:T_{\alpha }\to T_1\to T_2$, where ${\iota }_{\alpha }:T_{\alpha }\to T_1$ is the inclusion map, is continuous. And the disjoint union topology is the unique topology that makes the disjoint union topological space have that property.

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2: Proof

Let us suppose that $f\circ {\iota }_{\alpha }$ is continuous. For any open set, $U\subseteq T_2$, $(f\circ {\iota }_{\alpha }{)}^{-1}(U)={{\iota }_{\alpha }}^{-1}\circ f^{-1}(U)=f^{-1}(U)\cap T_{\alpha }$ is open. So, $f^{-1}(U)$ is open, by the definition of disjoint union topology. So, $f$ is continuous.

Let us suppose that $f$ is continuous. For any open set, $U\subseteq T_2$, $f^{-1}(U)$ is open. $f^{-1}(U)\cap T_{\alpha }$ is open, by the definition of disjoint union topology. $(f\circ {\iota }_{\alpha }{)}^{-1}(U)={{\iota }_{\alpha }}^{-1}\circ f^{-1}(U)=f^{-1}(U)\cap T_{\alpha }$, open. So, $f\circ {\iota }_{\alpha }$ is continuous.

Let us suppose that $T_1^{\prime }$ has the set of $T_1$ but has a not-necessarily disjoint union topology while it satisfies the property. Let us take $T_2$ to be $T_1^{\prime }$ and $f$ to be the identity map. Then, $f$ is continuous and $f\circ {\iota }_{\alpha }$ is continuous. For any open set, $U\subseteq T_1^{\prime }$, $(f\circ {\iota }_{\alpha }{)}^{-1}(U)=U\cap T_{\alpha }$ is open on $T_{\alpha }$. So, $T_2^{\prime }$ has the disjoint union topology or a smaller topology. Let us take $T_2$ to be $T_1$ (with the disjoint union topology) and $f$ to be the identity map. For any open set, $U\subseteq T_1$, $(f\circ {\iota }_{\alpha }{)}^{-1}(U)=U\cap T_{\alpha }$ is open on $T_{\alpha }$. So, $f\circ {\iota }_{\alpha }$ is continuous and $f$ is continuous. For any open set, $U\subseteq T_1$, $f^{-1}(U)=U$ is open on $T_1^{\prime }$. So, $T_1^{\prime }$ has no smaller topology after all.

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References

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