529: Restriction of Continuous Embedding on Domain and Codomain Is Continuous Embedding

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description/proof of that restriction of continuous embedding on domain and codomain is 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: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of continuous embedding.
• The reader knows a definition of subspace topology.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
• The reader admits the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets.

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Target Context

• The reader will have a description and a proof of the proposition that any restriction of any continuous embedding on the domain and the codomain is a continuous embedding.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous embeddings }\}$
$S_1$: $\subseteq T_1$
$S_2$: $\subseteq T_2$, $f(S_1)\subseteq S_2$
$f^{\prime }$: $:S_1\to S_2,p\mapsto f(p)$
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Statements:
$f^{\prime }\in \{\text{ the continuous embeddings }\}$
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2: Natural Language Description

For any topological spaces, $T_1,T_2$, any continuous embedding, $f:T_1\to T_2$, any subset, $S_1\subseteq T_1$, and any subset, $S_2\subseteq T_2$ such that $f(S_1)\subseteq S_2$, $f^{\prime }:S_1\to S_2,p\mapsto f(p)$ is a continuous embedding.

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3: Proof

$f^{\prime }$ is injective, because $f$ is so.

$f^{\prime }$ is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

Let us denote the codomain restriction of $f^{\prime }$ as $f^{″}:S_1\to f^{\prime }(S_1)$.

$f^{″}$ is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

As $f^{″}$ is bijective, there is the inverse, $f^{″-1}:f^{\prime }(S_1)\to S_1$. Is $f^{″-1}$ continuous?

For any open subset, $U\subseteq S_1$, is ${f^{″-1}}^{-1}(U)=f^{″}(U)$ open on $f^{\prime }(S_1)$?

$U=U^{\prime }\cap S_1$ for an open subset, $U^{\prime }\subseteq T_1$. $f^{″}(U)=f^{″}(U^{\prime }\cap S_1)=f^{″}(U^{\prime })\cap f^{″}(S_1)$, by the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets, $=f(U^{\prime })\cap f^{″}(S_1)=U^{″}\cap f(T_1)\cap f^{″}(S_1)$ where $U^{″}\subseteq T_2$ is an open subset, because $f:T_1\to f(T_1)$ is a homeomorphism (so, $f(U^{\prime })$ is open on $f(T_1)$), $=U^{″}\cap f^{″}(S_1)$, which is open on $f^{″}(S_1)=f^{\prime }(S_1)$.

So, yes, $f^{″-1}$ is continuous, and $f^{″}$ is a homeomorphism.

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References

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