2025-05-18

1116: For Open Surjective Continuous Map Between Topological Spaces, Image of Basis of Domain Is Basis of Codomain

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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


  • 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:
T1: { the topological spaces }
T2: { the topological spaces }
f: :T1T2, { the open surjective continuous maps }
J: { the possibly uncountable index sets }
B: { the bases of T1}, ={Bj|jJ}
B: =f(B):={f(Bj)|BjB}
//

Statements:
B{ the bases of T2}
//


2: Proof


Whole Strategy: Step 1: see that each f(Bj) is open on T2; Step 2: see that for each point, tT2, and each open neighborhood of t, Ut, there is an f(Bj) such that tf(Bj)Ut.

Step 1:

Each f(Bj) is open on T2 because f is open.

Step 2:

For any point, tT2, and any open neighborhood of t, UtT2, is there an f(Bj)B, such that tf(Bj)Ut?

f1(Ut) is open on T1 because f is continuous, and is not empty because f is surjective.

There is a point, tf1(Ut), such that f(t)=t, because f is surjective, and f1(Ut) is an open neighborhood of t.

There is a Bj such that tBjf1(Ut), because B is a basis.

t=f(t)f(Bj) and f(Bj)ff1(Ut)Ut, 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 T1 is 2nd-countable, T2 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(Bj) 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 T2 is not related with the topology of T1, so, there is no guarantee that there is a Bj such that f(Bj) 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.


References


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