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

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Topics

About: topological space

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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

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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.

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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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the open surjective continuous maps }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B$: $\in \{\text{ the bases of }{T}_1\}$, $=\{B_j|j\in J\}$
$B^{\prime }$: $=f(B):=\{f(B_j)|B_j\in B\}$
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Statements:
$B^{\prime }\in \{\text{ the bases of }{T}_2\}$
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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 $f(B_j)\in B^{\prime }$, 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, $t^{\prime }\in f^{-1}(U_t)$, such that $f(t^{\prime })=t$, because $f$ is surjective, and $f^{-1}(U_t)$ is an open neighborhood of $t^{\prime }$.

There is a $B_j$ such that $t^{\prime }\in B_j\subseteq f^{-1}(U_t)$, because $B$ is a basis.

$t=f(t^{\prime })\in f(B_j)$ 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.

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3: Note 1

So, if $T_1$ is 2nd-countable, $T_2$ is 2nd-countable, if $f$ satisfies the requirements of this proposition.

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4: Note 2

The requirements for $f$ are necessary for this proposition because if $f(B_j)$ is not open, $B^{\prime }$ cannot be any basis; if $f$ is not surjective, $B^{\prime }$ 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.

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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.

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References

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