Let $X$ be a topological space, with $\gamma : [0,1] \to X $ some path in $X$. Suppose $p : \hat{X} \to X$ is a covering map. Then, by the path-lifting lemma, we know that for each $\hat{x} \in p^{-1} ( \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to \hat{X}$ such that $\hat{\gamma}(0) = \hat{x}$.
Similarly, the homotopy lifting lemma asserts the existence of unique lifts for maps $I \times I \to X$.
It has previously been remarked that homotopies may be considered as paths in $C(X,Y)$ endowed with an appropriate topology (e.g. compact-open). Therefore, my question is:
Question: can the homotopy lifting lemma be deduced from the path-lifting lemma applied twice?
I offer an outline of a plausible proof, which I would struggle to verify as I am not well-acquainted with the compact-open topology (or any topology on $C(X,Y)$).
Let $H : I \times I \to X$ be our candidate map for lifting.
- Define $\gamma_i : [0,1] \to X, x \mapsto H(x,i)$ (for $i = 0,1$).
- These give unique lifts $\hat{\gamma_i} : [0,1] \to \hat{X}$.
- Consider $\gamma_i$ as elements of $C([0,1],X)$. Note that there is a path $\eta$ between them in this space, described by $H$.
- Prove that $C([0,1] ,\hat{X})$ is a covering space of $C([0,1],X)$.
- By the path-lifting lemma, there exists a unique lift of $\eta : [0,1] \to C([0,1],X)$ to $\hat{\eta}:[0,1] \to C([0,1],\hat{X})$.
- This implies a unique lift $\hat{H}:I \times I \to \hat{X}$ of $H$, concluding the proof.
The steps (1),(2) and (5) are consequences of the path-lifting lemma. (3) may be assumed given a suitable topology on $C(X,Y)$. Thus, this question boils down to proving that [there exists a topology on $C(X,Y)$ such that] steps (4) and (6) are valid.
Edit: a natural suggestion for the covering map in (4) is:
$$P : \hat{\gamma} \mapsto p \circ \hat{\gamma}$$
I am likely to have made a typographical error somewhere. Please do mention any you find.