description/proof of that codomain extension of open map into open subspace is open
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of open map.
- The reader knows a definition of topological subspace.
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that the codomain extension of any open map into any open subspace is open.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T'_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the open topological subspaces of } T'_2\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the open maps }\}\)
\(f'\): \(: T_1 \to T'_2, t_1 \mapsto f (t_1)\)
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Statements:
\(f' \in \{\text{ the open maps }\}\)
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2: Proof
Whole Strategy: Step 1: for each open subset, \(U_1 \subseteq T_1\), see that \(f' (U_1)\) is open on \(T'_2\).
Step 1:
Let \(U_1 \subseteq T_1\) be any open subset.
\(f' (U_1) = f (U_1) \subseteq T_2\) is open on \(T_2\).
\(f' (U_1) \subseteq T'_2\) is open on \(T'_2\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
3: Note
Compare with the proposition that a codomain extension of an open map is not necessarily open.
Compare with the proposition that any codomain restriction of any open map is open.
