Does a morphism between covering spaces define a covering? 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 $f$?
 A: If $p:Y\to X$ and $q:Z\to X$ are coverings of a locally connected topological space $X$, then any morphism  $f:Y\to Z $ of those coverings is a covering.
Proof
We may assume $X$ connected and $Y=X\times D, Z=X\times E$ both trivial ($D,E$ are discrete spaces).
The morphism $f$ is then of the form $f:   X\times D \to  X\times E: (x,d) \mapsto (x, \phi  (x,d))$.
The key point is that $\phi (x,d)$ actually doesn't depend on $x$ because $X$ is connected and $\phi  (x,d)\in E$, a discrete space.
Hence we may write $\phi  (x,d)=\Phi  (d)$ and we see that our  map $f$ is 
$$f: Y=X\times D \to Z= X\times E :(x,d)\mapsto (x,\Phi (d))$$
It is a covering because the inverse image of the open subset $X\times \lbrace e \rbrace \subset Z$ is $X \times \Phi^{-1} (e))$, a disjoint sum of copies of the open set, and because these open sets cover $Z$.
On surjectivity: a point of view
A covering space $f:Y\to Z $ is a continuous map such that $Z$ can be covered by open subsets $U$ over which $f$ is a trivial covering space $U\times D$ (where  $D$ is a discrete topological space).
This allows for the possibility that $D$ is empty, so that a covering space need not be surjective.
Of course  definitions are arbitrary and if you want you can demand that a covering be surjective.
I consider that a clumsy condition which will constantly force you to add to perfectly natural proofs  a surjectivity check. 
A: A recent paper 
Brazas, Jeremy, "Semicoverings: a generalization of covering space theory", Homology, Homotopy and Applications,  14 (2012) 33--63, (electronic)
gives a generalisation of covering maps, so that your $f: Y \to Z$ now becomes a semi-covering!
