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Let $f\colon X\to Y$ a continuous open and surjective function, where $X$ is Polish.

It is known that $Y$ is Polish if: $f$ is closed or $Y$ is metric.

Suppose that we know that $Y$ is Hausdorff, does it implies that $Y$ is Polish? or Regular?

I think no, but i have not been able to find a counterexample.

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2 Answers 2

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This seems to be Problem 327 of V. Tkachuk's book "A $C_p$ theory problem book: special features of function spaces", Springer, Problem books in Mathematics.

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    $\begingroup$ Yes but in that book (in page 27), the author says that all the spaces are Tychonoff, so $Y$ is regular. Since it is an open image of a Polish space, then $Y$ is second countable and thus metric by Uryshon. In this case $Y$ has to be Polish. $\endgroup$
    – Hector Pinedo
    Commented Oct 21, 2015 at 6:01
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As a counterexample one can consider the projective space $P\mathbb R^\omega$ of the countable product of lines. This is a quotient space of the Polish space $\mathbb R^\omega_\circ=\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$.

It is easy to see that the quotient map $q:\mathbb R^\omega_\circ\to P\mathbb R^\omega$ is open and the space $P\mathbb R^\omega$ is Hausdorff but not Urysohn and hence is not metrizable and not Polish.

Moreover, the space $P\mathbb R^\omega$ is superconnected in the sense that for any nonempty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection of thier closures $\bar U_1\cap\dots\cap\bar U_n$ is not empty.

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