Suppose $p:Y\to X$ is a covering map, $X,Y$ are manifolds and $X$ is connected. If $Z$ is a connected component of $Y$, I wonder if the restriction of $p$ on $Z$ is also a covering map? If not, what conditions should be added to guarantee the restriction is a covering map? (expect that $Y$ is compact.)
What I do: I know only need to show $p(Z)=X$. $p$ is a local homeomorphism and thus an open map. $Z$ is an open set since $Y$ is locally connected, and of course a closed set, so $p(Z)$ is an open set. Then I want to show $p(Z)$ is also closed, thus $p(Z)$ is open and closed in a connected space $X$, so $p(Z)=X$. But I cannot show $p(Z)$ is closed, perhaps I try a wrong way.
Best Answer
I'll use the definition of covering map as it appears in Hatcher's Algebraic Topology: A continuous map $p:Y\to X$ is called a covering map, if for every $x\in X$ there is an open neighborhood $U$ around $x$ whose preimage is a (possibly empty) disjoint union of open sets, each of which is mapped homeomorphically onto $U$ via $p$. Note that this definition does not require a covering map to be surjective.
Still, if the codomain is connected, the map must be surjective by the following argument:
Assume $x\notin p(Y)$. Then its preimage $p^{-1}(x)$ is empty. There is an open $U$ containing $x$ such that $p^{-1}(U)$ equals $\bigsqcup_{\alpha\in I}U_\alpha$ where $U_\alpha\approx U$ and all $U_\alpha$ are open. But since $x$ is not in the image of $p$, the disjoint union must be the empty union. This means that $U$ does not intersect $p(Y)$, so $p(Y)$ is closed.
On the other hand, $p$ is an open map. If $V\subset Y$ is an open set containing $y$ and $U$ is the evenly covered neighborhood of $p(y)$, then $y$ is contained in some $U_\alpha$. Since $V\cap U_\alpha$ is open, $p(V\cap U_\alpha)$ is an open subset of $U$, thus an open set in $X$ which is contained in $p(V)$, so $p(V)$ is open.
If $X$ is connected, then $p(Y)$ must be all of $X$, being a clopen subset of a connected space.
To prove that the restriction of the $p$ in your problem to the connected component $Z$ is also a covering map, take an $x\in X$ and an open neighborhood $U$ such that $p^{-1}(U)$ equals $\bigsqcup_{\alpha\in I}U_\alpha$ where $U_\alpha\stackrel p\approx U$ and all $U_\alpha$ are open. Since $X$ is locally connected, there is an open connected $V$ such that $x\in V\subset U$. Its preimage is $\bigsqcup_{\alpha\in I}V_\alpha$ where $V_\alpha$ is simply the preimage of $V$ in the particular $U_\alpha$. Each $V_\alpha$ is connected, so it is either entirely in $Z$ or it is disjoint to $Z$. If you delete all the $V_\alpha$'s which are not in $Z$ from the union, you obtain the preimage of $V$ under the restriction of $p$. This means that this restricted $p$ still is a covering map.