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Quotient map from $X$ to $Y$ is continuous and surjective with a property : $f^{-1}(U)$ is open in $X$ iff $U$ is open in $Y$.

But when it is open map? What condition need?

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    $\begingroup$ A quotient map $f \colon X \to Y$ is open if and only if for every open subset $U \subseteq X$ the set $f^{-1} (f (U))$ is open in $X$. A sufficient condition is that $f$ is the projection under a group action. $\endgroup$
    – Andrea
    Sep 1, 2011 at 13:40
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    $\begingroup$ @Andrea: "A sufficient condition is that f is the projection under a group action" Why, please? $\endgroup$
    – geometrian
    Oct 26, 2013 at 21:45
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    $\begingroup$ If $\pi \colon X \to X/G$ is the projection under the action of $G$ and $U \subseteq X$, then $\pi^{-1} (\pi (U)) = \cup_{g \in G} g(U)$. $\endgroup$
    – Andrea
    Oct 27, 2013 at 14:54
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    $\begingroup$ I'd like to add that the set $f^{-1}(f(U))$ described in Andrea's comment has a name. It's called the $f$-load of $U$. So a quotient map $f : X \to Y$ is open if and only if the $f$-load of every open subset of $X$ is an open subset of $X$. $\endgroup$
    – Matthew K.
    Oct 20, 2017 at 20:15
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    $\begingroup$ @MatthewK. It’s called the saturation. We may associate to any projection (or to any function) an equivalence relation: $x\sim y$ $:\Leftrightarrow$ $f(x)=f(y)$. Then $\begin{array}[t]{rcl} f^{-1}(f(U)) &= &\{x\mid f(x)\in f(U)\}\\ &= &\{x\mid\exists{y\in U:~}f(x)=f(y)\}\\ &= &\{x\mid\exists{y\in U:~}x\sim y\}\\ &= &\bigcup_{y\in U}[y]_{\sim},\\ \end{array}$ ie the saturation of $U$ under the equivalence $\sim$. $\endgroup$
    – Thomas
    Mar 9, 2019 at 14:32

2 Answers 2

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Let us agree that $f : X \to Y$ denotes a continuous surjection.

$f : X \to Y$ is called a quotient map if it satisfies the following condition:

$$U \subset Y \text{ is open } \Longleftrightarrow f^{-1}(U) \subset X \text{ is open.} \tag{Q}$$

The $\Longrightarrow$ part means nothing else than that $f$ is continuous which is redundant because $f$ was assumed to be continuous. We can therefore replace (Q) by

$$f^{-1}(U) \subset X \text{ is open } \Longrightarrow U \subset Y \text{ is open.} \tag{Q'}$$

We have the following well-known

Lemma. $f$ is a quotient map if and only it satisfies

$$f^{-1}(C) \subset X \text{ is closed } \Longrightarrow C \subset Y \text{ is closed.} \tag{Q''}$$

This gives the well-known

Lemma. If $f$ is an open or a closed map, then $f$ is a quotient map.

Proof. Let $f$ be an open map. If $f^{-1}(U)$ is open, then $f(f^{-1}(U))) = U$ is open. Thus $f$ is a quotient map. Note that the equation $f(f^{-1}(U))) = U$ requires surjectivity.
The case that $f$ is a closed map is treated similarly.

We have seen that being a quotient map is weaker than being an open map. Of course we get

Lemma. A quotient map $f$ is an open map if and only if $$U \subset X \text{ is open } \Longrightarrow f^{-1}(f(U)) \subset X \text{ is open.} \tag{O}$$

Proof. If $f$ is an open map, then $f(U)$ is open and thus $f^{-1}(f(U))$ is open. For the converse observe that $f^{-1}(f(U))$ open implies $f(U)$ open because $f$ is a quotient map.

There are certainly cases where the above lemma is very useful, but it depends on the concrete situation whether it is easier to verify condition (O) then to verify directly that $f$ is open.

The benefit of (O) is that we do not need to know the topology on $Y$ explicitly. This is particularly useful if we are given a surjection $f : X \to Y$ from a space $X$ to a set $Y$ and endow $Y$ with the quotient topology. If we are given a continuous surjection $f :X \to Y$ from a space $X$ to a space $Y$, it is probably easier to show directly that $f$ is an open map than proving first that $f$ is a quotient map and then verifying (O).

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  • $\begingroup$ In your 3rd lemma, why does $f^{-1}(f(U))$ open implies that $f(f^{-1}(f(U)))$ open? $\endgroup$ May 6, 2022 at 20:44
  • $\begingroup$ @CarsonJames See my update. $\endgroup$ May 8, 2022 at 10:06
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The definition of a quotient space:

A topological space $(Y,U)$ is called a quotient space of $(X,T)$ if there exists an equivalence relation $R$ on $X$ so that $(Y,U)$ is homeomorphic to $(X/R,T/R)$.

This definition is equivalent to:

$ (Y,U) $ is a quotient space of $(X,T)$ if and only if there exists a final surjective mapping $f: X \rightarrow Y$.

The previous statement says that $f$ should be final, which means that $U $ is the topology induced by the final structure,

$$ U = \{A \subset Y | f^{-1}(A) \in T \} $$

This is the largest collection that makes the mapping continuous, which is equivalently stated in your definition with the "if and only if" statement.

However, if the map is open:

If $f: X \rightarrow Y$ is a continuous open surjective map, then it is a quotient map.

Note that this also holds for closed maps. So in the case of open (or closed) the "if and only if" part is not necessary. If $f^{-1}(A)$ is open in $X$, then by using surjectivity of the map $f (f^{-1}(A))=A$ is open since the map is open. And the other side of the "if and only if" follows from continuity of the map.

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  • $\begingroup$ What is a final map? $\endgroup$ Oct 29, 2021 at 18:06

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