description/proof of that bijection between topological spaces that maps basis element to basis element and maps back basis element to basis element is homeomorphism
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of bijection.
- The reader knows a definition of basis of topological space.
- The reader knows a definition of homeomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any bijection between any topological spaces that maps each basis element to a basis element and maps back each basis element to a basis element is a homeomorphism.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(B_1\): \(\in \{\text{ the bases for } T_1\}\)
\(B_2\): \(\in \{\text{ the bases for } T_2\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the bijections }\}\)
//
Statements:
\(\forall b_1 \in B_1 (f (b_1) \in B_2) \land \forall b_2 \in B_2 (f^{-1} (b_2) \in B_1)\)
\(\implies\)
\(f \in \{\text{ the homeomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(t_1 \in T_1\) and any neighborhood of \(f (t_1)\), \(N_{f (t_1)} \subseteq T_2\), and choose an element of \(B_1\) that \(f\) maps into \(N_{f (t_1)}\); Step 2: see that the situation is symmetric for \(f^{-1}\).
Step 1:
Let \(t_1 \in T_1\) be any.
Let any neighborhood of \(f (t_1)\) be \(N_{f (t_1)} \subseteq T_2\).
There is a \(b_2 \in B_2\) such that \(f (t_1) \in b_2 \subseteq N_{f (t_1)}\), by the definition of basis of topological space.
\(f^{-1} (b_2) \in B_1\), by the supposition.
\(t_1 \in f^{-1} (b_2)\), because \(f (t_1) \in b_2\).
\(f^{-1} (b_2) \subseteq T_1\) is an open subset.
So, \(f^{-1} (b_2)\) is an open neighborhood of \(t_1\).
\(f (f^{-1} (b_2)) = b_2\), because \(f\) is a bijection, \(\subseteq N_{f (t_1)}\).
So, \(f\) is continuous at \(t_1\), but as \(t_1\) is arbitrary, \(f\) is continuous.
Step 2:
The situation is symmetric for \(f^{-1}\), so, \(f^{-1}\) is continuous.
