2

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 ...


1

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}(\...


Only top voted, non community-wiki answers of a minimum length are eligible