Is $\text{abs}: [-1, 1] \to [0,1]$ a covering map?

Question

I read that for any covering space $p:C \to X$ the cardinality of the fibre is the same for every $x \in X$ if $X$ is connected.

So what is wrong with my example: $\text{abs}: [-1,1] \to [0,1]$?

Clearly the fibre has cardinality 2 everywhere except for $x=0$ where the cardinality is 1. And I think this is a covering map since for all $x \in (0,1)$ $U = (0,1)$ is an open set satisfying the necessary conditions.

And for $x=0$ and $x=1$ we use $U = [0, 0.5)$ and $U = (0.5, 1]$ respectively.

So where am I wrong (or am I even wrong)?

Answer 1

$V=\operatorname{abs}^{-1}[0,\frac12)$ is a (path-)connected set and $\left.\operatorname{abs}\right\rvert_V:V\to [0,\frac12)$ is not a homeomorphism because it is not injective. Therefore $[0,\frac12)$ does not qualify as a valid trivialising neighbourhood of $0$.

Answer 2

$abs$ is not a covering map. That's because no open neighbourhood of $0$ in $[-1,1]$ can be mapped homeomorphically onto its image. And covering maps are local homeomorphisms.

In particular your $U=[0,0.5)$ example, as a neighbourhood of $0$ in the codomain, is not correct. Its preimage under $abs$ is $(-0.5,0.5)$ which cannot be decomposed into a disjoint union of open sets at all (it is connected). And so "being covering map" in that case means that $abs$ maps $(-0.5,0.5)$ homeomorphically onto $[0,0.5)$ which is clearly false.

However $abs$ is a covering map if we exclude $0$ from both domain and codomain. And in that case everything works.