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I asked this question on math@stackoverflow and was suggested to ask it here as well.

We call a surjective $f:X\rightarrow Y$ a quotient mapping if it satisfies, for every $U\subset Y$

(continuity, $f$ is continuous) $U$ is open $\Rightarrow$ $f^{-1}(U)$ is open

and

(???) $f^{-1}(U)$ is open $\Rightarrow$ $U$ is open.

I was wondering if there is a name for the second property itself of the definition, or should I just call it "the other" property?

And if a function satisfies this property only, how should I call it?

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  • $\begingroup$ What is (if any) the relation of (???) with the usual concept of open map? I mean, continuity + (???) imply that $f$ is open? $\endgroup$
    – Francesco Polizzi
    Commented Oct 21, 2014 at 21:05
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    $\begingroup$ I think it's not quite the same as openness: you do not require that the image of any open set should be open. Why do you need a name? It doesn't seem to make much sense without the continuity, and with continuity, you already know the name :) $\endgroup$
    – Alex Degtyarev
    Commented Oct 21, 2014 at 21:11
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    $\begingroup$ Sometimes it is said that $f$ is open with respect to saturated sets (i.e. sets of the form $f^{-1}(A)$). $\endgroup$
    – Ramiro de la Vega
    Commented Oct 22, 2014 at 0:59
  • $\begingroup$ @FrancescoPolizzi In this context openness (or closeness) is stronger in the sense that if a function is surjective continuous + open (or close) then it is a quotient map, but not vice versa. $\endgroup$
    – Vadim
    Commented Oct 22, 2014 at 3:18
  • $\begingroup$ @FrancescoPolizzi Consider $f:\mathbb{R}\rightarrow S^1$ given by $(\cos 2\pi x,\sin 2\pi x)$, this is a surjective, continuous and open map, so quotient, but it is not closed. Now restrict the same function to domain $[0,1]$, it is now surjective, continuous and closed map, so still quotient, but not open. An example from Munkres, the projection from $\mathbb{R}^2$ to $x$-axis restricted to domain $\mathbb{R}\times\{0\}\cup[0,+\infty)\times\mathbb{R}$ is a non-open non-closed quotient. $\endgroup$
    – Vadim
    Commented Oct 22, 2014 at 3:20

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