can we take skeletons of covering maps to give new covering maps? Let $X$ be an $n$-dimensional cell complex.
We attach an $(n+1)$-cell $e^{n+1}$ to $X$ and obtain a new cell complex $X'$.
Take the universal cover (or a general covering space) $\tilde X'$ of $X'$.
We have a covering map $\pi': \tilde X'\longrightarrow X'$.
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Question. Whether or not can we take the $n$-skeleton of $\pi'$ (restrict the map $\pi'$ to the $n$-skeleton of the CW complex $\tilde X'$)  and give a new covering map $\pi: \tilde X\longrightarrow X$?
Thanks for guidance.
 A: Yes, first one should check that the restriction of a covering space is a covering space. This is true either by just checking the axioms or appealing to the fact that the pullback of a covering space is a covering space, and the restriction of a covering space is just the pullback of an inclusion into the base space.
With that settled, the standard CW structure on the covering space induced by the CW structure on the base, we will have that $\tilde X'$ is obtained by attaching a number of (n+1)-cells, corresponding to the number of sheets of the covering, to the standard CW structure on $\tilde X$, the total space of the restriction of this cover to $X$. Hence, under this CW structure restriction to the n-skeleton, has codomain $X$, so this gives a cover of $X$.
Just because I find juggling all these X's confusing I will be very explicit, what we have is a map of pairs $(\tilde X',\tilde X) \rightarrow (X',X)$ where the map of the larger spaces is a covering space, and its restriction is one as well, and both of these pairs are a CW complex and its n-skeleton.
Supposing the cover is connected and $n\geq 1$, the subgroup of $\pi_1(X)$ classifying this cover will be the preimage of the subgroup of $\pi_1(\tilde X)$ classifying $\tilde X'$ under the induced map $\pi_1(X) \rightarrow \pi_1(X')$.
