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Let $(X,A)$ and $(Y,B)$ pairs satisfying the homotopy extension property (or a CW-pairs if necessary) with $A$ and $B$ contractible. Let $f:(X,A)\to (Y,B)$ a map of pairs and let $p_X:\widetilde{X}\to X$ and $p_Y:\widetilde{Y}\to Y$ covering maps.

By definition of covering space, every point $x\in X$ has a neighbourhood $U$ such that $p^{-1}(U)$ is a disjoint union of open sets each of them mapped homeomorphically onto $U$ by $p$, so in particular $p^{-1}(x)$ is a discrete set with one point on each component of $p^{-1}(U)$. By the theory of covering spaces we know that under certain conditions, a map $(X,x_0)\to (Y,y_0)$ may lift to $(\widetilde{X},\tilde{x}_0)\to (\widetilde{Y},\tilde{y}_0)$.

I am interested in knowing if such property are satisfied in my setting, namely, is $p^{-1}(A)$ a disjoint union of subspaces of $\widetilde{X}$ each of them mapped homeomorphically by $p$ onto $A$.

If so, I can choose $\widetilde{A}$ and $\widetilde{B}$ to be such subspaces of $\widetilde{X}$ and $\widetilde{Y}$ respectively. Then I wonder if the map $f:(X,A)\to (Y,B)$ can be (uniquely) lifted to $\tilde{f}:(\widetilde{X},\widetilde{A})\to (\widetilde{Y}, \widetilde{B})$ making the diagram of maps of pairs and covering maps commute.

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In general a fibre bundle over a contractible space is trivial, so $p_X^{-1}(A)\cong A \times F_{x_0}$ where $F_{x_0}$ is the (in our case discrete) fibre over any point in $A$.

Wether or not a map $f\colon (X, A) \to (Y, B)$ lifts to $\tilde{f}\colon(\tilde{X}, \tilde{A}) \to (\tilde{Y}, \tilde{B})$ then doesn't really depend on $A$ or $B$ because they are homotopically trivial, so it's essentially the same lifting problem as if they were points.

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Isn't it automatic ? (edit : I've actually assumed that in the decomposition $p_Y^{-1}(B) = \displaystyle\coprod_{i\in I}B$, each of the $B$ was a connected component; and that the coverings were normal)

Indeed, consider the basepoints $x_0,y_0$ to be in $\tilde{A},\tilde{B}$ respectively; and consider a lift $g:(\tilde{X},\tilde{x_0})\to (\tilde{Y},\tilde{y_0})$.

Then $g(\tilde{A})$ is a connected subset of $\tilde{Y}$ that contains $\tilde{y_0}$. Moreover, $p_Y(g(\tilde{A}))=f(p_X(\tilde{A})) = f(A)\subset B$. So $g(\tilde{A})$ is included in $p_Y^{-1}(B)$, more specifically in the connected component containing $\tilde{y_0}$, which is $\tilde{B}$; so $f$ is actually a map of pairs.

Now if, say the baspoint $\tilde{y_0}$ is not in $\tilde{B}$, but in some other connected component of $p_Y^{-1}(B)$, then up to a deck transformation, you have the same result. So as long as the basepoints $x_0,y_0$ are in $A,B$ respectively, you can find such a lift. Now if $X,Y$ are path-connected, you can assume that this is the case.

So it suffices that the map can be lifted to a $g$ for it to be lifted to a map of pairs

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