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As Moishe Kohan explained in his comments, covering projections are fiber bundles with discrete fibers, at least if the base space $B$ is connected (If it is not, we can write $B$ as the union of nonempty disjoint open subspaces $U_1, U_2$ and there exist coverings with fibers of different cardinality over $U_1, U_2$. This would no longer be a fiber bundle ...
Well, let me be the first to tell you that this is a very special case of a pullback, in this case of covers (but more generally bundles.) Hint: take a point $y \in Y$, and consider $f(y)$. Take a neighborhood $U$ around it so that there is a homeomorphism $p^{-1}(U) \cong U \times F$, and consider the preimage of this under $f$, i.e: $f^{-1}(U)$. Let $\pi:... 1 You simply have to check that the$U'_\alpha$are the path components of$p^{-1}(U')$. First of all, they are disjoint and open by (1) and the choices made. Moreover,$U'$is path-connected and$U'_\alpha$is homeomorphic to it under$p$, so$U_\alpha'$is also path-connected. The rest follows simply. 1 That will not be enough : indeed the assumption on$\pi_1(Y,y_0)$only tells us about the path-component of$y_0$, nothing else. To get a specific counterexample, take$X\to B$to be the$2$-sheeted connected covering of$S^1$(so$z\mapsto z^2, S^1\to S^1$),$b_0 = 1, x_0 = 1$; and take$Y = S^1\sqcup \mathbb R$,$y_0 = 0 \in \mathbb R$and$f: Y\to B$... 1 I think Wolfgang Globke's answer could be generalised in the following way. Fix a point$x_0\in X$. The cover$\pi\colon Y\to X$corresponds to a subgroup$H\subset \pi_1(Y,x_0)$. A loop$\gamma$in$Y$based in a point$y_0\in\pi^{-1}(x_0)$is null-homotopic if and only if$\pi(\gamma)$lies in$H$. Define$\$\mathrm{Hol}^H(X,x_0)=\{\mathrm{hol}_{x_0}(\...