# Properties of covering spaces replacing base points by contractible subspaces

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.

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.

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