376: Characteristic Property of Subspace Topology

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A description/proof of characteristic property of subspace topology

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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 subspace 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 subspace of any topological space, any map from any topological space to the subspace is continuous if and only if the composition of the inclusion of the subspace to the superspace after the map is continuous, and the subspace topology is the unique topology that makes the subspace 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 topological space, $T_2$, and any subspace, $T_3\subseteq T_2$, any map from any topological space, $f:T_1\to T_3$, is continuous if and only if $\iota \circ f:T_1\to T_3\to T_2$, where $\iota :T_3\to T_2$ is the inclusion map, is continuous. And the subspace topology is the unique topology that makes the subspace have that property.

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2: Proof

Let us suppose that $\iota \circ f$ is continuous. For any open set, $U\subseteq T_3$, $U=U^{\prime }\cap T_3$ where $U^{\prime }\subseteq T_2$ is open on $T_2$ by the definition of subspace topology. $f^{-1}(U)=(\iota \circ f{)}^{-1}(U^{\prime })$, because $(\iota \circ f{)}^{-1}(U^{\prime })=f^{-1}\circ {\iota }^{-1}(U^{\prime })=f^{-1}(U^{\prime }\cap T_3)=f^{-1}(U)$. So, $f^{-1}(U)$ is open, so, $f$ is continuous.

Let us suppose that $f$ is continuous. For any open set, $U^{\prime }\subseteq T_2$, $(\iota \circ f{)}^{-1}(U^{\prime })=f^{-1}\circ {\iota }^{-1}(U^{\prime })=f^{-1}(U^{\prime }\cap T_3)$. So, $(\iota \circ f{)}^{-1}(U^{\prime })$ is open, so, $\iota \circ f$ is continuous.

Let us suppose that $T_3^{\prime }$ has the set of $T_3$ but has a not-necessarily subspace topology while it satisfies the property. Let us take $T_1$ to be $T_3^{\prime }$ and $f$ to be the identity map. Then, $f$ is continuous and $\iota \circ f$ is continuous. For any open set, $U^{\prime }\subseteq T_2$, $(\iota \circ f{)}^{-1}(U^{\prime })=U^{\prime }\cap T_3$ is open on $T_3^{\prime }$. So, $T_3^{\prime }$ has the subspace topology or a larger topology. Let us take $T_1$ to be $T_3$ (with the subspace topology) and $f$ to be the identity map. For any $U^{\prime }\subseteq T_2$, $(\iota \circ f{)}^{-1}(U^{\prime })=U^{\prime }\cap T_3$, open on $T_3$. So, $\iota \circ f$ is continuous and $f$ is continuous. For any open set, $U\subseteq T_3^{\prime }$, $f^{-1}(U)=U$ is open on $T_3$. So, $T_3^{\prime }$ has no larger topology after all.

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References

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