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Suppose $M$ is a connected topological manifold, which has a finite open cover $\{U_i\}$, where each $U_i$ is homeomorphic to $\mathbb{R}^n$. Is it necessarily true that $\pi_1(M)$ is finitely generated?

I tried to adapt the proof when $M$ is compact, but there are still subtleties I can't seem to get around; mainly, what happens in an intersection $U_i\cap U_j$. In the compact case you can refine the cover first to nicer sets, then take a finite subcover. But here that doesn't seem possible.

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Consider $\Bbb R^2$ with a countable infinite discrete set removed, say the points $(n,0)$ for integer $n$. This can be covered by two sets homeomorphic to $\Bbb R^2$. But $\pi_1(M)$ is not finitely generated. A finite number of closed paths in $M$ will fail to wind about $(n,o)$ for $n$ large enough.

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  • $\begingroup$ Wow, thanks! On a related note, do you know if every manifold can be finitely covered by such sets? $\endgroup$ – Steve D Feb 10 '18 at 7:41

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