description/proof of that local homeomorphism is open
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of local homeomorphism.
- The reader knows a definition of open map.
- 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.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any local homeomorphism 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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the local homeomorphisms }\}\)
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Statements:
\(f \in \{\text{ the open maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any open subset \(U \subseteq T_1\), for each \(u \in U\), take a homeomorphic restriction, \(f \vert_{U_u}: U_u \to U_{f (u)}\), and see that \(f \vert_{U_u} (U_u \cap U)\) is open on \(T_2\); Step 2: see that \(f (U)\) is open on \(T_2\).
We will use 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 without mentioned further.
Step 1:
Let \(U \subseteq T_1\) be any open subset of \(T_1\).
For any point, \(u \in U\), there are an open neighborhood of \(u\), \(U_u \subseteq T_1\), and an open neighborhood of \(f (u)\), \(U_{f (u)} \subseteq T_2\), such that \(f \vert_{U_u}: U_u \to U_{f (u)}\) is a homeomorphism.
\(U_u \cap U\) is open on \(U_u\), and so, \(f \vert_{U_u} (U_u \cap U) \subseteq U_{f (u)}\) is open on \(U_{f (u)}\), and is open on \(T_2\).
Step 2:
\(f (U) = f (\cup_{u \in U} U_u \cap U) = \cup_{u \in U} f (U_u \cap U)\), 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, \(= \cup_{u \in U} f \vert_{U_u} (U_u \cap U)\), which is open on \(T_2\) as a union of open subsets.
