description/proof of that for open surjective continuous map between topological spaces, image of basis of domain is basis of codomain
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Proof
- 3: Note 1
- 4: Note 2
- 5: Note 3
Starting Context
- The reader knows a definition of continuous map.
- The reader knows a definition of open map.
- The reader knows a definition of surjection.
- The reader knows a definition of basis of topological space.
- The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
Target Context
- The reader will have a description and a proof of the proposition that for any open surjective continuous map between any topological spaces, the image of any basis of the domain is a basis of 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(f\): \(: T_1 \to T_2\), \(\in \{\text{ the open surjective continuous maps }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(\in \{\text{ the bases of } T_1\}\), \(= \{B_j \vert j \in J\}\)
\(B'\): \(= f (B) := \{f (B_j) \vert B_j \in B\}\)
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Statements:
\(B' \in \{\text{ the bases of } T_2\}\)
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2: Proof
Whole Strategy: Step 1: see that each \(f (B_j)\) is open on \(T_2\); Step 2: see that for each point, \(t \in T_2\), and each open neighborhood of \(t\), \(U_t\), there is an \(f (B_j)\) such that \(t \in f (B_j) \subseteq U_t\).
Step 1:
Each \(f (B_j)\) is open on \(T_2\) because \(f\) is open.
Step 2:
For any point, \(t \in T_2\), and any open neighborhood of \(t\), \(U_t \subseteq T_2\), is there an \(f (B_j) \in B'\), such that \(t \in f (B_j) \subseteq U_t\)?
\(f^{-1} (U_t)\) is open on \(T_1\) because \(f\) is continuous, and is not empty because \(f\) is surjective.
There is a point, \(t' \in f^{-1} (U_t)\), such that \(f (t') = t\), because \(f\) is surjective, and \(f^{-1} (U_t)\) is an open neighborhood of \(t'\).
There is a \(B_j\) such that \(t' \in B_j \subseteq f^{-1} (U_t)\), because \(B\) is a basis.
\(t = f (t') \in f (B_j)\) and \(f (B_j) \subseteq f \circ f^{-1} (U_t) \subseteq U_t\), by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
So, yes.
3: Note 1
So, if \(T_1\) is 2nd-countable, \(T_2\) is 2nd-countable, if \(f\) satisfies the requirements of this proposition.
4: Note 2
The requirements for \(f\) are necessary for this proposition because if \(f (B_j)\) is not open, \(B'\) cannot be any basis; if \(f\) is not surjective, \(B'\) cannot cover the area into which \(f\) does not map; if \(f\) is not continuous, the openness of a subset of \(T_2\) is not related with the topology of \(T_1\), so, there is no guarantee that there is a \(B_j\) such that \(f (B_j)\) is contained in the subset.
5: Note 3
As a quotient map is not necessarily open, this proposition cannot be applied to general quotient maps, so, a quotient space of a 2nd-countable topological space is not guaranteed to be 2nd-countable by this proposition, and in fact, there are some non-2nd-countable quotient spaces of 2nd-countable topological spaces.
