A description/proof of that injective map between topological spaces is continuous embedding if domain restriction of map on each element of open cover is continuous embedding onto open subset of range or codomain
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous embedding.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- 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 open set on any open topological subspace is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that any injective map between any topological spaces is a continuous embedding if the domain restriction of the map on each element of any open cover is any continuous embedding onto any open subset of the range or the codomain.
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: Description
For any topological spaces, \(T_1, T_2\), any injective map, \(f: T_1 \to T_2\), and any open cover, \(\{U_\alpha \vert \alpha \in A\}\), of \(T_1\) where \(A\) is any possibly uncountable indices set, \(f\) is a continuous embedding if each \(f \vert_{U_\alpha}: U_\alpha \to T_2\) is any continuous embedding onto any open subset of \(f (T_1)\) or \(T_2\).
When each \(f \vert_{U_\alpha}\) is onto any open subset of \(T_2\), \(f\) is onto a open subset of \(T_2\).
2: Proof
If \(f \vert_{U_\alpha} (U_\alpha)\) is open on \(T_2\), it will be open on \(f (T_1)\), because \(f \vert_{U_\alpha} (U_\alpha) = f \vert_{U_\alpha} (U_\alpha) \cap f (T_1)\).
So, let us suppose that \(f \vert_{U_\alpha} (U_\alpha)\) is open on \(f (T_1)\).
\(f\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
Let us define \(f': T_1 \to f (T_1)\), \(p \mapsto f (p)\).
\(f'\) is continuous, the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
As \(f\) is injective, \(f'\) is injective and there is the inverse, \(f'^{-1}: f (T_1) \to T_1\). Is \(f'^{-1}\) continuous?
Let us \(U \subseteq T_1\) be any open subset. \({f'^{-1}}^{-1} (U) = f' (U) = f' (U \cap \cup_{\alpha \in A} U_\alpha) = f' (\cup_{\alpha \in A} (U \cap U_\alpha)) = \cup_{\alpha \in A} f' (U \cap U_\alpha)\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets. But as \(f \vert_{U_\alpha}\) is a continuous embedding and \(U \cap U_\alpha \subseteq U_\alpha\) is open on \(U_\alpha\), \(f' (U \cap U_\alpha) \subseteq f (U_\alpha)\) is open on \(f (U_\alpha)\), but as \(f (U_\alpha) \subseteq T_2\) is open on \(f (T_1)\), \(f' (U \cap U_\alpha)\) is open on \(f (T_1)\), by the proposition that any open set on any open topological subspace is open on the base space. So, \(f' (U) = \cup_{\alpha \in A} f' (U \cap U_\alpha) \subseteq f (T_1)\) is open on \(f (T_1)\).
So, yes, \(f'^{-1}\) is continuous.
So, \(f'\) is a homeomorphism, and so, \(f\) is a continuous embedding.
When \(f \vert_{U_\alpha} (U_\alpha) \subseteq T_2\) is open on \(T_2\), \(f (T_1) = f (\cup_{\alpha \in A} U_\alpha) = \cup_{\alpha \in A} f (U_\alpha)\), which is open on \(T_2\).
3: Note
The requirement that each \(f \vert_{U_\alpha}\) is onto any open subset of \(f (T_1)\) is necessary for \(f\)'s being guaranteed to be a continuous embedding, unless proved otherwise.
