1
$\begingroup$

Let $Y$ be a locally compact $\sigma$-compact Hausdorff space and $q:Y\to X$ a quotient map with $X$ Hausdorff. Consider the following properties.

I. There is a dense subset $D\subseteq Y$ such that the restriction of $q$ to $D$ is injective.

II. The interior of every fibre (i.e. set of the form $q^{-1}(x)$, $x\in X$) in $Y$ is either empty or a singleton.

Then clearly I.$\Rightarrow$II. Suppose that $Y$ is a separable metric space satisfying II. Let $D_0$ be the (countable) set of isolated points in $Y$ and fix a countable base $\{U_i\}_{i\ge 1}$ for $Y\setminus \overline{D_0}$. Inductively chose a point from each $U_i$ which does not belong to any fibre with non-empty interior or any fibre previously chosen. This is possible because $U_i$ contains no isolated point and hence is an uncountable Baire space, and every fibre intersects $U_i$ in a closed set with empty interior. Let $D_1$ be the set thus obtained, and set $D=D_0\cup D_1$. Then $D$ is dense in $Y$ and $q$ restricted to $D$ is injective. So for locally compact separable metric spaces, II.$\Rightarrow$I.

Question: Does II.$\Rightarrow$I. for all locally compact $\sigma$-compact Hausdorff spaces?

$\endgroup$
1
$\begingroup$

We have II and not I when $X$ is any non-empty locally compact $\sigma$-compact Hausdorff space without isolated points, $Z$ is any compact Hausdorff space such that the density of $Z$ is bigger than $|X|$, $Y=X\times Z$, and $q:X\times Z\to X$ is the projection map.

$\endgroup$
1
  • 1
    $\begingroup$ Very good! Many thanks once again. $\endgroup$ – user558840 Oct 3 '20 at 7:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.