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Let's say that a smooth manifold $M$ is $k$-coverable if there exists a surjective smooth submersion $N \to M$ such that $H^k(N, \mathbb{R}) = 0$.

For example, every manifold is $1$-coverable by the universal cover.

What about $k = 2, 3, ...$? (I'm not assuming compactness of either $M$ or $N$.)

More generally, if $\omega$ is a closed $k$-form on $M$, is there a surjective submersion $f : N \to M$ (possibily depending on $\omega$) such that $f^*\omega$ is exact?

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    $\begingroup$ I think one may take $N = \Bbb R^{\dim M}$: choose a metric on $M$ with injectivity radius is bounded below by $\epsilon > 0$, choose a smooth immersion $\gamma: \Bbb R \to M$ so that every point of $M$ lies within distance $\epsilon/2$ to a point in the image of the immersion, choose a trivialization of the normal bundle to $\gamma$, and then consider the map $B_\epsilon(\Bbb R^{n-1}) \times \Bbb R \to M$ given by $(v, t) \mapsto \exp_{\gamma(t)}(v)$ (here identifying $v$ with an element of the normal space $N_{\gamma(t)} M$.) This map or some variation should be a surjective local diffeo. $\endgroup$
    – mme
    Commented Dec 15, 2023 at 14:43
  • $\begingroup$ @mme Many thanks for your comment. Do we need compactness to find a metric with an injectivity radius bounded below? $\endgroup$
    – Mattis Bakken
    Commented Dec 15, 2023 at 14:49
  • $\begingroup$ No (the term you want is 'bounded geometry', see the references here: mathoverflow.net/q/244469/40804). But I would expect the construction I outlined isn't really in the spirit of what you're looking for? $\endgroup$
    – mme
    Commented Dec 15, 2023 at 14:53
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    $\begingroup$ I should check the details in some of the claims above, which I will try to do in a few days and post an answer (if anyone else has a reference or can do the details quickly, they should go ahead). $\endgroup$
    – mme
    Commented Dec 15, 2023 at 14:59
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    $\begingroup$ Since $N$ is not required to be connected in the statement, there is a much easier answer: pick any atlas N→M, where N is a disjoint union of charts in M, which we can take to be diffeomorphic to open balls. Now N is homotopy equivalent to a discrete space (with H^k=0 for k>0) and any closed form on N is exact. $\endgroup$
    – Dmitri Pavlov
    Commented Dec 15, 2023 at 18:03

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