1117: Map Between Topological Spaces Is Open if but Not Only if Map Is Surjective and Open Subset of Domain Is Preimage of Open Subset of Codomain

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description/proof of that map between topological spaces is open if but not only if map is surjective and open subset of domain is preimage of open subset 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: Note
• 3: Proof
• 4: Note

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Starting Context

• The reader knows a definition of open map.
• The reader knows a definition of Euclidean topology.
• The reader admits the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset.

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Target Context

• The reader will have a description and a proof of the proposition that any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset 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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the maps }\}$
//

Statements:
(
$f\in \{\text{ the surjections }\}$
$\wedge$
$\mathrm{\forall }{U}_1\in \{\text{ the open subsets of }{T}_1\}(\mathrm{\exists }{U}_2\in \{\text{ the open subsets of }{T}_2\}(U_1=f^{-1}(U_2)))$
)
$⟹$
$f\in \{\text{ the open maps }\}$
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The reverse does not necessarily hold.

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2: Note

$f$ does not need to be continuous.

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3: Proof

Whole Strategy: Step 1: suppose that $f$ is surjective and $U_1=f^{-1}(U_2)$, and see that $f$ is open; Step 2: see an example in which the reverse does not hold.

Step 1:

Let us suppose that $f$ is surjective and any open subset, $U_1\subseteq T_1$, is $U_1=f^{-1}(U_2)$ for an open subset, $U_2\subseteq T_2$.

$f(U_1)=f\circ f^{-1}(U_2)$, but $f\circ f^{-1}(U_2)=U_2$, by the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset. So, $f(U_1)=U_2$, open on $T_2$.

Step 2:

Let us see an example in which the reverse does not hold.

Let $T_1$ be $\mathbb{R}$ with the Euclidean topology, $T_2$ be $\{0\}$ with the inevitable topology, $f$ be the constant map. $f$ is open. $f$ is surjective, but the open subset, $(0,1)$, is not the preimage of any open subset of $T_2$.

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4: Note

The surjectivity is required. As a counterexample, let $T_1$ be $(0,1]\subseteq \mathbb{R}$ with the subspace topology of the Euclidean topological space, $T_2$ be $\mathbb{R}$ with the Euclidean topology, $f$ be the identity map. For any open subset, $U_1\subseteq T_1$, $U_1=U_1^{\prime }\cap (0,1]$ where $U_1^{\prime }\subseteq \mathbb{R}$ is open, by the definition of subspace topology, and $U_1=f^{-1}(U_1^{\prime })$. But while $(1/2,1]\subseteq T_1$ is open, $f((1/2,1])=(1/2,1]$ is not open on $T_2$. That is because $f$ is not surjective.

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References

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