description/proof of that expansion of continuous embedding on codomain is continuous embedding
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
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.
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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T'_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\subseteq T'_2\) with the subspace topology
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous embeddings }\}\)
\(S\): \(\subseteq T'_2\) with the subspace topology such that \(T_2 \subseteq S\)
\(f'\): \(: T_1 \to S, p \mapsto f (p)\)
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Statements:
\(f' \in \{\text{ the continuous embeddings }\}\)
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2: Natural Language Description
For any topological spaces, \(T_1, T'_2\), any subspace, \(T_2 \subseteq T'_2\), any continuous embedding, \(f: T_1 \to T_2\), and any subset, \(S \subseteq T'_2\) with the subspace topology such that \(T_2 \subseteq S\), \(f': T_1 \to S, p \mapsto f (p)\) is a continuous embedding.
3: Proof
Whole Strategy: Step 1: see that \(f'\) is a continuous injection; Step 2: see that the codomain restriction of \(f'\), \(f'' : T_1 \to f' (T_1)\), is a homeomorphism.
Step 1:
\(f'\) is injective, because \(f\) is so.
\(f'\) 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 \(f'\) as \(f'' : T_1 \to f' (T_1)\).
\(f''\) 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, \(f''^{-1}: f' (T_1) \to T_1\). Is \(f''^{-1}\) continuous?
Set-wise, \(f' (T_1) = f (T_1)\). \(f' (T_1)\) as the subspace of \(S\) 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. \(f' (T_1)\) as the subspace of \(T'_2\) 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 \(f''^{-1}: f' (T_1) \to T_1\) is nothing but the inverse for the original \(f\), which is continuous, because \(f\) is a continuous embedding.
So, yes, \(f''^{-1}\) is continuous.
