description/proof of that restriction of continuous embedding on domain and 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.
- 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.
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.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous embeddings }\}\)
\(S_1\): \(\subseteq T_1\)
\(S_2\): \(\subseteq T_2\), \(f (S_1) \subseteq S_2\)
\(f'\): \(: S_1 \to S_2, p \mapsto f (p)\)
//
Statements:
\(f' \in \{\text{ the continuous embeddings }\}\)
//
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 \(f (S_1) \subseteq S_2\), \(f': S_1 \to S_2, p \mapsto f (p)\) is a continuous embedding.
3: Proof
\(f'\) is injective, because \(f\) is so.
\(f'\) 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 \(f'\) as \(f'' : S_1 \to f' (S_1)\).
\(f''\) 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, \(f''^{-1}: f' (S_1) \to S_1\). Is \(f''^{-1}\) continuous?
For any open subset, \(U \subseteq S_1\), is \({f''^{-1}}^{-1} (U) = f'' (U)\) open on \(f' (S_1)\)?
\(U = U' \cap S_1\) for an open subset, \(U' \subseteq T_1\). \(f'' (U) = f'' (U' \cap S_1) = f'' (U') \cap f'' (S_1)\), 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, \(= f (U') \cap f'' (S_1) = U'' \cap f (T_1) \cap f'' (S_1)\) where \(U'' \subseteq T_2\) is an open subset, because \(f: T_1 \to f (T_1)\) is a homeomorphism (so, \(f (U')\) is open on \(f (T_1)\)), \(= U'' \cap f'' (S_1)\), which is open on \(f'' (S_1) = f' (S_1)\).
So, yes, \(f''^{-1}\) is continuous, and \(f''\) is a homeomorphism.
