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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.

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    $\begingroup$ 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. $\endgroup$ – Thomas Rot Mar 30 '20 at 14:29
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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.

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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)

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