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Let $M$ be a smooth manifold and $p:M \to N$ a covering map. Then $N$ is a topological manifold but is it necessarily a smooth manifold? If yes, does there exist a smooth structure on $N$ making $p$ into a submersion?

If $\Gamma$ is a discrete group acting freely and properly on $M$ by diffeomorphisms then of course $M/\Gamma$ can be made smooth such that $M \to M/\Gamma$ is a submersion. My question (equivalent to it anyways) is: what if $\Gamma$ only acts on $M$ by homeomorphisms?

Constructing a counterexample may be a bit tedious. The "simplest" topological manifold that isn't a smooth manifold I am aware of doesn't have any non trivial covers ($E_8$ is simply connected).

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    $\begingroup$ (This example is simpler than $E_8$) $\endgroup$
    – Didier
    Nov 9, 2023 at 18:39

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The answer is negative. For instance, in section 2 of

Ruberman, Daniel, Invariant knots of free involutions of S^4, Topology Appl. 18, 217-224 (1984). ZBL0559.57016.

Ruberman constructs a non-smoothable fake $RP^4$: It is the quotient of $S^4$ by a fixed-point free involution. (Ruberman notes that one can also prove the existence of such manifold using Freedman's work.)

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