Suppose I have the following theorem (1):
If $C,X$ are spaces, $p:C\to X$ is a covering map, $Y$ is a "nice" topological space (I think simply connected and locally path-connected is sufficient), and $c_0\in C$, $y_0\in Y$, then for any continuous function $f:Y\to X$ there is a unique "lift" $\hat f:Y\to C$ such that $\hat f\circ p=f$ and $\hat f(y_0)=c_0$.
I want to extend it to the following (2):
If $C,X,Z$ are spaces, $p:C\to X$ is a covering map, $Y$ is a "nice" topological space, $c:Z\to C$ is continuous, and $y_0\in Y$, then for any continuous function $f:Y\times Z\to X$ there is a unique "lift" $\hat f:Y\times Z\to C$ such that $\hat f\circ p=f$ and $\forall z\in Z,\hat f(y_0,z)=c(z)$.
Since (1) had a fairly difficult proof, I am hoping that I can prove (2) as a corollary of (1). There are two obvious approaches:
Choose the spaces $C,X$ in (1) to be $C^Z,X^Z$, probably with the compact-open topology. The problem with this is that a lot of the exponential laws don't seem to hold without additional assumptions on $Z$, which I'd like to avoid. Furthermore, is it true that the promoted covering map $p':C^Z\to X^Z$ defined by $p'(f)(z)=p(f(z))$ is also a covering map?
Use (1) separately for each $z$, producing a family of continuous functions $\hat {f_z}:Y\to C$. While it is certainly true that the full lift if it exists must satisfy $\hat f(y,z)=\hat{f_z}(y)$, it is not at all obvious to me that the composite function defined this way is continuous.
Any suggestions for the more fruitful approach? Or are both doomed to fail because extra assumptions on $Z$ are necessary?
