Restriction of Covering Spaces to a component. Let $p:X\rightarrow Y$ be a covering space (i.e. locally over $Y$, $p$ looks like the projection $U\times F\rightarrow U$, with $F$ discrete).
If $Y$ is locally connected, it is known that the restriction of $p$ to a connected component of $X$ is also a covering space.
I was trying to find a counterexample to this fact when $Y$ is connected but not locally connected. Any idea?
 A: Ok, I think I find an example which comes from the theory of solenoids (see https://mathoverflow.net/questions/219911/does-every-connected-component-of-a-covering-space-over-a-connected-base-interse). Let me spell it down, maybe it is useful for someone else who does not know solenoids.
Take $X=\mathbb{R}\times \mathbb{Z}_2$. Then $\mathbb{Z}$ acts on $X$ topologically free, by
$n\cdot (x,y) = (x+n,y-n)$. It follows that the quotient
$X\rightarrow X/\mathbb{Z}$ is a covering space.
The space $\mathbb{Z}_2$ is totally disconnected, so the connected components of $X$ are all of the form $\mathbb{R} \times \{y\}$, for some $y\in \mathbb{Z}_2$.
Notice that the restriction $\mathbb{R} \times \{y\}\rightarrow X/\mathbb{Z}$ is not surjective and so it is not a covering map.
However, the quotient $X/\mathbb{Z}$ is connected!  ($X/\mathbb{Z}$ has also other nice properties: it is a metrizable compact topological groups, but we do not need this here)
If $U$ is a $\mathbb{Z}$-saturated clopen in $X$, then $U=\mathbb{R}\times Y$ for some $Y\subset \mathbb{Z}_2$, which is in turn a $\mathbb{Z}$-saturated clopen in $\mathbb{Z}_2$. We claim that either $Y=\emptyset$ or $Y=\mathbb{Z}_2$.
In fact, for $y \in Y$ and $z\in \mathbb{Z}_2$, for every $k\in \mathbb{N}$ we can find $n\in \mathbb{Z}$ such that $p^k|(y-z-n)$. So the $\mathbb{Z}$-orbit of $z$ intersect every open neighborhood of $y$, hence $z\in Y$.
