If the fibers of a quotient map are all discrete, is this map a covering map?

Question

If $p:\tilde{X}\rightarrow X$ is a covering projection then I know that for every point $x \in X$ the fibre above $x$, i.e $p^{-1}(x)$, has the discrete topology. Here $p$ being a covering map means that it is a continuous, surjective, and every $x \in X$ has a neighborhood $U$ such that $p^{-1}(U)$ is homeomorphic to a disjoint union of copies of $U$, each of which mapped homeomorphically to $U$ through $p$.

Is the converse true? I.e, if $\forall x\in X$ if $p^{-1}(x)$ has the discrete topology and p is also a quotient map then is it true that $p$ is a covering projection? Or is there a counter example for the same?

I am not sure how to proceed in either direction. Hints will be appreciated.

Thanks!

Answer 1

No, that's not true. Consider $p : \mathbb R \to [0, +\infty)$, $x \mapsto x^2$. Every fiber is a finite discrete set, and it is a quotient map: it's surjective, continuous, and $U \subset [0, +\infty)$ is open iff its preimage is open:

• Suppose $U$ is open, and let $x \in p^{-1}(U)$. Then $x^2 \in U$, so there's a small neighborhood of $x^2$ contained in $U$. If $x = 0$ this small neighborhood can be taken to be $[0, \epsilon)$, and then $(-\sqrt{\epsilon}, \sqrt{\epsilon}$ is a small neighborhood of $0$ contained in $p^{-1}(U)$. Otherwise the preimage of an open interval around $x^2$ contains an open interval around $x$. In both cases $p^{-1}(U)$ contains a neighborhood of $x$. This was true for any $x$, so $p^{-1}(U)$ is open.
• Suppose $p^{-1}(U)$ is open and let $x \in U$, $x = y^2$ for $y \in U$. $p^{-1}(U)$ contains a small interval around $y$, and the image through $p$ of this interval is again an interval, containing $y^2 = x$. Again by looking at both cases ($x = 0$, $x \neq 0$) you can prove this is either an interval of the type $[0,\epsilon)$ or an open interval, both of which are open in $[0, +\infty)$. Thus $U$ is open.

But this isn't a covering map, because otherwise the fiber would have constant cardinality, which isn't the case.

It's really necessary to have the condition that the fiber "stays the same over a neighborhood" (and varies continuously), and this is precisely what the definition of a covering map encodes.