description/proof of that codomain extension of open map is not necessarily 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.
Target Context
- The reader will have a description and a proof of the proposition that a codomain extension of an open map is not necessarily 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T'_2\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological subspaces of } T'_2\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the open maps }\}\)
\(f'\): \(: T_1 \to T'_2, t_1 \mapsto f (t_1)\)
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Statements:
not necessarily \(f' \in \{\text{ the open maps }\}\)
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2: Proof
Whole Strategy: Step 1: see a counterexample.
Step 1:
Let \(T_1 = [0, 1) \subseteq \mathbb{R}\) as the topological subspace of the Euclidean topological space, \(\mathbb{R}\), \(T'_2 = \mathbb{R}\), the Euclidean topological space, \(T_2 = [0, 1) \subseteq T'_2\), and \(f = id\).
\(f\) is open, because it is the identity map.
But \(f'\) is not open, because while \([0, 1) \subseteq T_1\) is open, \(f' ([0, 1)) = [0, 1) \subseteq T'_2\) is not open on \(T'_2\).
3: Note
Compare with the proposition that the codomain extension of any open map into any open subspace is open.
Compare with the proposition that any codomain restriction of any open map is open.
