description/proof of that continuous open or closed injection between topological spaces is continuous embedding
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous embedding.
- The reader knows a definition of open map.
- The reader knows a definition of closed map.
- The reader knows a definition of injection.
- 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 any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
- The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
Target Context
- The reader will have a description and a proof of the proposition that any continuous open or closed injection between any topological spaces 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the continuous injections }\}\)
//
Statements:
\(f \in \{\text{ the open maps }\} \cup \{\text{ the closed maps }\}\)
\(\implies\)
\(f \in \{\text{ the continuous embeddings }\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f\) is open, and see that \(f\) is a continuous embedding; Step 2: suppose that \(f\) is closed, and see that \(f\) is a continuous embedding.
Step 1:
Let us suppose that \(f\) is open.
Let us think of the codomain restriction of \(f\), \(f': T_1 \to f (T_1), t \mapsto f (t)\).
\(f'\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
As \(f\) is an injection, \(f'\) is a bijection, and so, there is the inverse, \(f'^{-1}: f (T_1) \to T_1\).
Let \(U \subseteq T_1\) be any open subset.
\({f'^{-1}}^{-1} (U) = f' (U) = f (U) \subseteq T_2\) is open, because \(f\) is open.
\(f' (U) = f' (U) \cap f (U)\), because \(f' (U) \subseteq f (U)\) anyway, and \(f' (U)\) is open on \(f (U)\), by the definition of subspace topology.
So, \(f'^{-1}\) is continuous, by Note for the definition of continuous, topological spaces map.
So, \(f'\) is a homeomorphism.
So, \(f\) is a continuous embedding.
Step 2:
Let us suppose that \(f\) is closed.
Let us think of the codomain restriction of \(f\), \(f': T_1 \to f (T_1), t \mapsto f (t)\).
\(f'\) is continuous, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
As \(f\) is an injection, \(f'\) is a bijection, and so, there is the inverse, \(f'^{-1}: f (T_1) \to T_1\).
Let \(U \subseteq T_1\) be any closed subset.
\({f'^{-1}}^{-1} (U) = f' (U) = f (U) \subseteq T_2\) is closed, because \(f\) is closed.
\(f' (U) = f' (U) \cap f (U)\), because \(f' (U) \subseteq f (U)\) anyway, and \(f' (U)\) is closed on \(f (U)\), by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.
So, \(f'^{-1}\) is continuous, by the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
So, \(f'\) is a homeomorphism.
So, \(f\) is a continuous embedding.
