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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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    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.2011-09-01
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    anything that is locally a projection should be open2011-09-01
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    Another sufficient condition: that the map be a surjective submersion of smooth manifolds.2013-05-17
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    @Andrea: "A sufficient condition is that f is the projection under a group action" Why, please?2013-10-26
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    If $\pi \colon X \to X/G$ is the projection under the action of $G$ and $U \subseteq X$, then $\pi^{-1} (\pi (U)) = \cup_{g \in G} g(U)$.2013-10-27
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    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 $X$.2017-10-20

1 Answers 1

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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.