Noncompact surface with finite fundamental group

Question

Let $M$ be a noncompact surface with finite fundamental group. Can we prove that $M$ must be homeomorphic to $\mathbb R^2$?

Answer 1

Let $M$ be a non-compact manifold with finite fundamental group. The universal cover of $M$ is either $S^2$ or $\mathbb{R}^2$, but $M$ is non-compact, so it must be the latter; in particular, $M$ is aspherical. As the fundamental group of an aspherical manifold is torsion-free (see below), we see that $M$ is simply connected and hence homeomorphic to its universal cover, namely $\mathbb{R}^2$.

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Proposition: Let $X$ be a finite-dimensional aspherical CW complex. Then $\pi_1(X)$ is torsion-free.

Proof: If $\alpha$ is a torsion element of $\pi_1(X)$, then $\langle \alpha\rangle \cong \mathbb{Z}_k$ for some $k$; let $X' \to X$ be the corresponding covering. Then $X'$ is a finite-dimensional aspherical CW complex with $\pi_1(X') \cong \mathbb{Z}_k$, so $X'$ is a $K(\mathbb{Z}_k, 1)$. As Eilenberg-MacLane spaces are unique up to homotopy, we see that $X'$ is homotopy equivalent to the infinite-dimensional lens space $S^{\infty}/\mathbb{Z}_k$. Therefore $H^n(X'; \mathbb{Z}_k) \cong H^n(S^{\infty}/\mathbb{Z}_k; \mathbb{Z}_k) \cong \mathbb{Z}_k$ for every $n$, but this is impossible as $X'$ is finite-dimensional.