926: Expansion of Continuous Embedding on Codomain Is Continuous Embedding

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description/proof of that expansion of continuous embedding on 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 of subset of topological space.
• The reader admits the proposition that any expansion of any continuous map on the codomain is continuous.
• The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that any expansion of any continuous embedding on 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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\subseteq T_2^{\prime }$ with the subspace topology
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous embeddings }\}$
$S$: $\subseteq T_2^{\prime }$ with the subspace topology such that $T_2\subseteq S$
$f^{\prime }$: $:T_1\to S,p\mapsto f(p)$
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Statements:
$f^{\prime }\in \{\text{ the continuous embeddings }\}$
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2: Natural Language Description

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

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3: Proof

Whole Strategy: Step 1: see that $f^{\prime }$ is a continuous injection; Step 2: see that the codomain restriction of $f^{\prime }$, $f^{″}:T_1\to f^{\prime }(T_1)$, is a homeomorphism.

Step 1:

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

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

Step 2:

Let us denote the codomain restriction of $f^{\prime }$ as $f^{″}:T_1\to f^{\prime }(T_1)$.

$f^{″}$ 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, $f^{″-1}:f^{\prime }(T_1)\to T_1$. Is $f^{″-1}$ continuous?

Set-wise, $f^{\prime }(T_1)=f(T_1)$. $f^{\prime }(T_1)$ as the subspace of $S$ is the subspace of $T_2^{\prime }$, by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace. $f^{\prime }(T_1)$ as the subspace of $T_2^{\prime }$ is the subspace of $T_2$, by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace. That means that $f^{″-1}:f^{\prime }(T_1)\to T_1$ is nothing but the inverse for the original $f$, which is continuous, because $f$ is a continuous embedding.

So, yes, $f^{″-1}$ is continuous.

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References

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