I am considering $k$-fold covers of smooth manifolds (with smooth covering maps).

Let $f:M^m\to N^m$ be a smooth finite covering map.


The following implication is not true: $M$ can be embedded into $\mathbb{R}^n$ $\Rightarrow$ $N$ can be embedded into $\mathbb{R}^n$.

The easiest counterexample is $M=\mathbb{S}^{n-1}$ and $N=RP^{n-1}$.


But what about the converse of the implication:

$N$ can be embedded into $\mathbb{R}^n$ $\Rightarrow$ $M$ can be embedded into $\mathbb{R}^n$.

Intuitively, this makes sense since if $N$ can be embedded, we should be able to embed $M$ which is obtained by "cutting, unfolding and gluing several copies of $N$" and is therefore "less complicated".

Or is there an obvious counterexample I'm missing?

NB: I'm of course aware of the Whitney Embedding Theorem. The point is, however, to derive something better than $n=2m$ for $M$ based on the fact that I know something about the embeddability of $N$.


The converse is also false. This is because there are planar surfaces that are finitely covered by non-planar surfaces. The first example that comes to mind is the four-holed torus double covering the four-holed sphere (three-holed disk).

In dimension three there are many more examples - for example most finite covers of knot complements do not embed in euclidean three-space.

[Edit] It appears that adding the hypothesis of "closed" doesn't save us -- Section Three of (http://arxiv.org/abs/0810.2346) gives a few examples of closed three-manifolds (of Nil and Solv geometry) that embed in $S^4$ and which have (infinitely many) finite covers that do not embed in $S^4$.

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    $\begingroup$ Thanks, that helped! Actually (and I should've mentioned that above) I am mostly interested in manifolds with boundary, so that's all the counterexamples I needed at the moment. But if a compact-without-boundary example comes to your head, let me know! $\endgroup$ – Fabian Jun 9 '15 at 17:40

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