Let $$\pi:X \rightarrow Y$$ be a surmersion (surjective submersion) between closed manifolds.

1) Is there any obstruction to the existence of a "multi-valued" section $$s$$ of $$\pi$$ such that $$\pi \circ s$$ is a smooth covering of $$Y$$ ?

By a multi-valued section I was thinking about gluing local section of $$\pi$$, where by a local section I mean a map $$\sigma : U \rightarrow X$$ with $$U$$ and open subset of $$Y$$ and satisfying $$\pi \circ \sigma = id_U$$. The multi-valued section should take the form of an immersion $$s : Y' \rightarrow X$$ with $$p: Y' \rightarrow Y$$ is a covering map and such that $$\pi \circ s = p$$.

2) Is it always possible to restrict us to finite covering $$p : Y' \rightarrow Y$$ ?

A trivial example is given when the fiber bundle is trivial and the covering is just the trivial covering.

• If $Y$ is simply connected there are no covers, then the question reduced to the question if a section always exists right? That is not the case. For example the unit sphere bundle $X=T_1S^2$ over $Y=S^2$ does not have a section. – Thomas Rot Mar 30 '20 at 14:29

First, since you are dealing with closed manifolds, a submersion $$\pi:X\rightarrow Y$$ (assuming that the base is connected) is surjective because it is simultaneously open and closed. Moreover, due to Ehresmann's Theorem, the submersion $$\pi:X\rightarrow Y$$ is a fiber bundle as it is proper.

Here is an approach for constructing non-examples: Let $$Y$$ be a compact manifold with $$\pi_1(Y)$$ finite. Denote its universal cover by $$p:\tilde{Y}\rightarrow Y$$. Pulling back the fiber bundle $$\pi:X\rightarrow Y$$ with the finite cover $$p:\tilde{Y}\rightarrow Y$$ results in a new fiber bundle $$\tilde{X}\rightarrow\tilde{Y}$$. It is not hard to see that the natural map $$\tilde{X}\rightarrow X$$ is also a finite cover. Hence a multi-valued section $$s:\tilde{Y}\rightarrow X$$ lifts to a section of the bundle $$\tilde{X}\rightarrow\tilde{Y}$$ because $$\pi_1(\tilde{Y})$$ is trivial. So we need to arrange for the pullback of the bundle $$\pi:X\rightarrow Y$$ with the universal covering map $$p:\tilde{Y}\rightarrow Y$$ to do not admit a section.

1) Similar to @ThomasRot's comment, you may consider sphere bundles on $$Y$$: Let $$E\rightarrow Y$$ be an oriented vector bundle of rank $$r$$ whose Euler class $$e(E)\in H^{r}(Y,\Bbb{R})$$ is non-zero. Now take $$X$$ to be the sphere bundle $$S(E)\rightarrow Y$$. If it admits a multi-valued section, from the discussion above the same must be true for its pullback $$S(\tilde{E})\rightarrow \tilde{Y}$$ where $$\tilde{E}\rightarrow\tilde{Y}$$ is the pullback of the bundle $$E\rightarrow Y$$ with the finite (universal) cover $$p:\tilde{Y}\rightarrow Y$$. But the former bundle cannot admit a section since the Euler class of the vector bundle $$\tilde{E}\rightarrow\tilde{Y}$$, given by $$p^*(e)$$, is non-zero because $$p^*:H^{r}(Y,\Bbb{R})\rightarrow H^{r}(\tilde{Y},\Bbb{R})$$ is injective.

2) Another approach is to take $$\pi:X\rightarrow Y$$ to be a principal $$G$$-bundle. Recall that such bundles are trivial if they admit a section, and their isomorphism classes are in bijection with homotopy classes of maps from the base to the classifying space $$BG$$. Thus any map $$Y\rightarrow BG$$ with the property that the composition $$\tilde{Y}\rightarrow Y\rightarrow BG$$ is not null-homotopic provides you with a principal $$G$$-bundle $$\pi:X\rightarrow Y$$ without any multi-valued section.

Here is a simple counterexample. Let $$X=(0,2/3)\cup(4/3,2)$$ (open intervals), $$Y=(0,1)$$ and $$\pi(x)=x$$ if $$x\in(0,2/3)$$ and $$\pi(x)=x-1$$ if $$x\in(4/3,2)$$. Note that $$Y$$ being simply connected has no nontrivial cover.

If you add the condition that the fibres are connected, then you will have counterexamples with higher-dimensional $$Y$$.

Maybe you can have a look at a paper of mine: Submersions, fibrations and bundles, Trans. Amer. Math. Soc. 354 (2002), 3771-3787

(avaible here: http://web.univ-ubs.fr/lmba/meigniez/docu/travaux.html)