My question involves topological spaces $X$, $Y$ and $Z$, two coverings $p : Y \rightarrow X$ and $q: Z \rightarrow X$ of $X$ and a morphism $f: Y \rightarrow Z$ of coverings, i.e. a map which fulfills $p = q \circ f$.
My question is: Is $f$ itself a covering map $Y \rightarrow Z$ ?
First, I thought it should be, because given an element $z \in Z$ we have an open neighborhood $U \subseteq X$ with $q(z) \in U$ such that $p^{-1}(U) = \bigcup_{i \in I}U_i$ and $q^{-1}(U) = \bigcup_{j \in J} V_j$ are disjoint unions of open subsets with $p|_{U_i} : U_i \rightarrow U$ and $q|_{V_j} : V_j \rightarrow U$ homeomorphims. Because of $p = q \circ f$ I get $ \bigcup_{i \in I}U_i = p^{-1}(U) = f^{-1}(q^{-1}(U)) = \bigcup_{j \in J} f^{-1}(V_j).$ Now there is exactly one $j \in J$ with $z \in V_j$ and if I let I':= \{i \in I \:|\: f(U_i) \subseteq V_j\} for all i \in I' I have $f|_{U_i} = (q|_{V_j})^{-1} \circ p|_{U_i}$ (right?) which then would be a homeomorphism. But now I still have to show something like f^{-1}(V_j) = \bigcup_{i \in I'} U_i to get a disjoint union of opens which are mapped homeomorphically onto $V_j$ and this is where I started to doubt my first assessment. I am not able to show why it is impossible for some $i \in I$ that $f(U_i) \cap V_j \neq \emptyset$ without $f(U_i) \subseteq V_j$. I believe if I assumed $X$ to be locally connected then I could choose the $U_i$ to be connected and I should be done.
So, in fact there are two things I would like to know, namely if my argument up to the point I stated it is valid and if one can go on and show the "proposition" without assuming local connectedness.
EDIT: As was pointed out, the answer to my question is negative if $f$ is not surjective. Is there a chance of saving it by assuming surjectivity and continuity of $f$?