395: Universal Property of Continuous Embedding

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A description/proof of universal property of continuous embedding

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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 embedding.
• The reader knows a definition of subspace topology.

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Target Context

• The reader will have a description and a proof of the universal property of continuous embedding: any injection between topological spaces is a continuous embedding if and only if any additional map from any additional topological space into the domain of the original map is continuous if and only if the composition of the additional map before the original map 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 injection, $f_1:T_1\to T_2$, $f_1$ is a continuous embedding if and only if for any topological space, $T_3$, and any map, $f_2:T_3\to T_1$, $f_2$ is continuous if and only if $f_1\circ f_2:T_3\to T_2$ is continuous.

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2: Proof

Suppose that $f_1$ is a continuous embedding. Suppose $f_2$ is continuous. Then, $f_1\circ f_2$ is continuous as a compound of continuous maps. Suppose $f_1\circ f_2$ is continuous. Then, for any open set, $U\subseteq T_1$, $f_1(U)$ is open on $f_1(T_1)$ while $f_1(U)=U^{\prime }\cap f_1(T_1)$ where $U^{\prime }\subseteq T_2$ is open on $T_2$ by the definition of subspace topology. As $f_1\circ f_2$ is continuous, $(f_1\circ f_2{)}^{-1}(U^{\prime })=f_2^{-1}\circ f_1^{-1}(U^{\prime })$ is open on $T_3$. But $f_1^{-1}(U^{\prime })=f_1^{-1}(U^{\prime }\cap f_1(T_1))=f_1^{-1}(f_1(U))$, but $f_1^{-1}(f_1(U))=U$, because $f_1$ is bijective. So, $f_2^{-1}(U)$ is open on $T_3$, which means that $f_2$ is continuous.

Suppose that for any topological space, $T_3$, and any map, $f_2:T_3\to T_1$, $f_2$ is continuous if and only if $f_1\circ f_2:T_3\to T_2$ is continuous. Let us take $T_3=T_1$ and $f_2:T_1\to T_1$ as the identity map, continuous. So, $f_1\circ f_2=f_1:T_1\to T_2$ is continuous. As $f_1$ is injective, $f_1^{\prime }:T_1\to f_1(T_1)$ is bijective, so, let us take $T_3:=f_1(T_1)$ where $T_3$ is the subspace topological space and $f_2:={f_1^{\prime }}^{-1}:f_1(T_1)\to T_1$. $f_1\circ f_2:f_1(T_1)\to T_2$ is an inclusion, so, continuous. So, $f_2={f_1^{\prime }}^{-1}$ is continuous, so, $f_1^{\prime }$ is a homeomorphism.

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References

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