When is a quotient map open?

Question

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?

Answer 1

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).