Let $X$ be a metric space and let $\pi:X\to Y$ be an open, continuous, surjective map (quotient map) onto locally compact Hausdorff space $Y$ (in particular, $Y$ is completely regular). Does it follow that $Y$ is a normal space?
I don't know which assumptions are needed and if it is true. (It can be assumed moreover: $X$ has finite covering dimension, i.e. Lebesgue dimension. It can be also helpful for me to show that $Y$ has finite covering dimension in this setting).
EDIT: If $X$ is separable then it is secong countable and since the quotient map is open, $Y$ is second countable as well. Now, $Y$ is a second countable, regular Hausdorff space and by Urysohn's metrization theorem it is metrizable. In this case, we are done.
Thank you for your time and for any help!
There is a fairly easy zero-dimensional counterexample. I have no proof that it is strongly zero-dimensional, but that seems quite likely.
The counterexample I am referring to is the rational sequence topology . It is shown in many places that this is a zero-dimensional locally compact Hausdorff space, but not normal.
In addition, since basic open sets are either singletons or homeomorphic to the one-point compactification of $\mathbb{N}$, it is locally metrizable. The latter property implies that it is an image of a metrizable space by a open continuous mapping. Indeed, if $C$ is a cover by metrizable open subspaces, then $X = \coprod_{U\in C} U$ is metrizable and the combination of inclusion mappings $U\hookrightarrow Y$ is a local homeomorphism.