Let $M$ be a connected topological 2-manifold. I.e. a Hausdorff space locally homeomorphic to $\mathbb{R}^2$. Suppose $\pi_1(M)$ is a finitely generated free group. Must $M$ be homeomorphic to a closed surface with finitely many punctures?

I don't think it matters, but I actually only care about orientable $M$.

This is inspired by the answer to this question: Can two different topological spaces cover each other?

  • $\begingroup$ You should add 2nd countable or paracompactness assumption, otherwise, the claim is false already in the simply connected case (product of two long lines). $\endgroup$ – Moishe Kohan Sep 10 '16 at 11:21

Yes. An open surface is either a finitely punctured surface or one has infinitely generated homology groups. A very light sketch is to pick an exhaustion by compact connected submanifolds; at each stage you're either gluing on cylinders, capping off boundary components with discs (or in the nonorientable case Mobius bands), or increasing the genus. If your homology is finitely generated, then your genus is finite, so eventually you're only adding on punctured spheres or capping off components; then show that adding punctured spheres without capping stuff off increases the rank of the homology, bounding the number of ends you can have.

A more precise argument in the case of $\pi_1 = 0$ is given here. You can modify the argument for the case of finitely-generated $H_1(\Sigma;\Bbb Z/2)$.

There is a classification of open surfaces following similar ideas: they are classified by their genus, whether or not they're orientable, and some data about precisely how they're noncompact: their space of ends, their space of non-orientable ends, and their space of ends with genus. If the fundamental group is finitely generated, it can't have any non-orientable ends or ends with genus; these contribute an infinitely generated summand to first homology. (This is why it's nice to work with homology instead of fundamental groups: Having an infinitely generated subgroup is ok, but not if you're abelian!)

At this point, we know that our surface is of finite genus, but it still might have a complicated space of ends (each end being genus zero). Each end contributes to homology: a manifold with at least $n$ ends has homology of rank at least $n-1$. So you had better have finitely many. Thus, you're a finitely punctured closed surface.

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    $\begingroup$ @Timkinsella That's a fair complaint, I do assume second countable. But I believe there are already uncountably many simply connected surfaces if you don't assume they're second countable. As a single example, you can take the universal cover of the Prufer surface. $\endgroup$ – user98602 Aug 22 '16 at 22:20
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    $\begingroup$ @Timkinsella No need; since $\pi_1$ is countable for any second-countable manifold, so is every subgroup, and thus the beginning of your third sentence applies. Any "reasonable", everyday operation you can do to the class of second-countable manifolds stays within the class of second-countable manifolds. $\endgroup$ – user98602 Aug 22 '16 at 23:21
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    $\begingroup$ Mike: The classification of open surfaces is more complicated than you stated (i.e. either a tame surface or a surface of infinite genus). See my answer here: math.stackexchange.com/questions/459160/…. For instance, consider the complement to the Cantor set in $S^2$. Of course, the conclusion still holds since $H_1$ is of finite rank. $\endgroup$ – Moishe Kohan Aug 23 '16 at 10:30
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    $\begingroup$ @MikeMiller my comment I'd about the first sentence of your answer $\endgroup$ – Moishe Kohan Aug 23 '16 at 16:35
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    $\begingroup$ @studiosus Oh, as you say that's obviously wrong. I don't know why that's there. Thank you. $\endgroup$ – user98602 Aug 25 '16 at 20:12

It is a bit annoying that, while it is well-known, there does not seem to be a complete proof of this result anywhere, while most pieces of the proof are in the literature. Mike Miller's answer is also missing details.

In what follows, an $n$-dimensional manifold is a 2nd countable Hausdorff space locally homeomorphic to $R^n$. (In particular, by default, manifolds are not assumed to have boundary.)

I will start with a general definition:

Definition. A topological $n$-manifold $M$ is said to be tame if there exists a compact $n$-manifold with boundary $\bar{M}$ such that $M$ is homeomorphic to the interior of $\bar{M}$, i.e. $\bar{M} -\partial \bar{M}$.

In this case, the inclusion $M\to \bar{M}$ is a homotopy-equivalence. In particular (if $M$ is connected), $\pi_1(M)\cong \pi_1(\bar{M})$ is finitely-presented. Thus, every connected tame manifold has finitely presented fundamental group. The converse is famously false already in dimension 3: The Whitehead manifold is contractible but is not tame. In what follows, will be using homology with $Z_2$-coefficients.

The situation in the case of surfaces, on the other hand, is much simpler:

Theorem 1. Suppose that $M$ is a noncompact connected surface. Then the following are equivalent:

  1. $M$ is tame.

  2. $H_1(M):=H_1(M; Z_2)$ is finitely generated (i.e. is finite).

  3. $\pi_1(M)$ is finitely-generated.

  4. $\pi_1(M)$ is finitely-presented.

  5. $\pi_1(M)$ is free of finite rank.

Proof. The implications (5)$\Rightarrow$(4)$\Rightarrow$(3)$\Rightarrow$(2) are immediate (the last one uses the Hurewicz theorem). The implication (1)$\Rightarrow$(5) follows from: If $M$ is tame then $\pi_1(M)\cong \pi_1(\bar{M})$. Since $M$ is assumed to be noncompact, $\bar{M}$ has nonempty boundary, hence, by applying inductively the van Kampen theorem, we conclude that $\pi_1(\bar{M})$ is free. Alternatively, one can use the much harder result that fundamental groups of connected noncompact surfaces are free.

Thus, it remains to prove the implication (2)$\Rightarrow$(1).

Since all surfaces admit triangulations, we can assume that $M$ is triangulated. Let $M^1$ denote the 1-skeleton of the triangulation. Then $H_1(M^1)$ maps onto $H_1(M)$. Since $H_1(M)$ is assumed to be finitely generated and $M$ is connected, there is a finite connected subgraph $X\subset M^1$ such that the inclusion maps $X\to M^1\to M$ induce a surjective map $H_1(X)\to H_1(M)$. Let $Y\subset M$ be the regular neighborhood of $X$ in $M$. It is a compact surface with boundary. The inclusion $X\to Y$ is also a homotopy-equivalence, hence, the natural homomorphism $H_1(Y)\to H_1(M)$ is also surjective. Consider a boundary component $C$ of $Y$. If $C$ does not separate $M$ then, applying the Mayer-Vietoris sequence (or duality), we conclude that $H_1(X)\to H_1(M)$ is not surjective, which is impossible. Thus, $C$ separates $M$ in two components, one of which contains the interior of $X$ and the other, $X_C$, is disjoint from $X$. If the inclusion map $C\to \bar{X}_C$ (the closure of $X_C$ in $M$) does not induce a surjection $H_1(C)\to H_1(\bar{X}_C)\cong H_1(X_C)$, then applying the Mayer-Vietoris sequence, we again see that $H_1(X)\to H_1(M)$ is not surjective, which is a contradiction. In particular, by the classification of compact surfaces, either $\bar{X}_C$ is homeomorphic to a closed disk, or $X_C$ is noncompact and, thus, $H_1(C)\to H_1(X_C)$ is 1-1.

We, therefore, enlarge $X$ by adding to it the complementary components $X_C$ homeomorphic to $R^2$. The new subsurface with boundary, denoted $Z$, still has the property that the inclusion $Z\to M$ induces a surjective homomorphism $H_1(Z)\to H_1(M)$ (since $Z$ contains $X$). Now, however, the homomorphism $H_1(Z)\to H_1(M)$ is also injective. Therefore, $H_1(Z)\to H_1(M)$ is an isomorphism. Let $N$ denote the interior of $Z$: It is a noncompact tame surface ($Z=\bar{N}$). We still have isomorphisms induced by inclusions $$ H_1(N)\cong H_1(Z)\cong H_1(M). $$

So far, the proof was self-contained. I will now use:

Theorem 2. Suppose that $S'\subset S$ is a connected subsurface $S'$ in a noncompact connected surface $S$ such that the inclusion $S'\to S$ induces an isomorphism $H_1(S')\to H_1(S)$. Then $S'$ is homeomorphic to $S$.

See Theorem 7.1 in

Martin Goldman, An algebraic classification of noncompact 2-manifolds. Trans. Amer. Math. Soc. 156 (1971), 241–258.

This theorem implies that the surfaces $N$ and $M$ are homeomorphic. Since $N$ is tame, so is $M$. qed

Remark. Since fundamental groups of noncompact connected surfaces are free, finiteness of $H_1(M; Z_2)$ is equivalent to the finite generation of $\pi_1(M)$. This allows for an alternative geometric proof of the implication (2)$\Rightarrow$(1) which I give below. (But the proof uses much heavier machinery.)

Equip $M$ with an arbitrary Riemannian metric. This, in turn, defines a conformal structure on $M$. Lift this conformal structure to the universal covering $\tilde{M}\to M$. By the Uniformization Theorem, $\tilde{M}$ is conformal to either the open unit disk or to the complex plane. Accordingly, we obtain a properly discontinuous, free, isometric action of $\Gamma=\pi_1(M)$ on the hyperbolic plane ${\mathbb H}^2$ (in the unit disk model) or the Euclidean plane $E^2$. I will consider the former case since the latter is easier. We have a discrete finitely generated subgroup $\Gamma< Isom({\mathbb H}^2)$. Every such subgroup admits a finitely-sided fundamental polygon $F\subset {\mathbb H}^2$, see for instance Alan Beardon's book "The Geometry of Discrete Groups."

The sides of $F$ are pairwise identified by some generators $\gamma_1,...,\gamma_n$ of $\Gamma$ and the quotient of $F$ by this identification is homeomorphic to our surface $M$. The polygon $F$ is noncompact (since $M$ is). Its noncompactness stems from two possible sources:

(a) $F$ may have "ideal vertices", points in the (ideal) boundary circle of ${\mathbb H}^2$, where two of the sides of $F$ meet.

(b) $F$ may have "ideal edges", arcs in the boundary of ${\mathbb H}^2$.

We enlarge $F$ by adding to it the ideal vertices and edges, the result is $\bar{F}$, a compact polygon. Identifying its boundary edges results in a compact surface with boundary $S$: The boundary of $S$ is the image of the union of ideal edges of $F$. The (finitely many) ideal vertices of $F$ project to finitely many points $p_1,...,p_k$ in $S- \partial S$. Thus, $M$ is homeomorphic to the complement $$ S- (\partial S \cup \{p_1,...,p_k\}). $$
Equivalently, let $D_1,...D_k$ denote small (pairwise disjoint and disjoint from $\partial S$) closed disk neighborhoods of $p_1,...,p_k$. Then $M$ is homeomorphic to the complement $$ S- (\partial S \cup D_1\cup ...\cup D_k). $$
Let $\bar{M}$ denote $M- (cl(D_1)\cup ...\cup cl(D_k))$. Then $\bar{M}$ is a compact surface with boundary and $M$ is homeomorphic to the interior of $M$. qed

Edit. Here is one more proof of the implication (3)$\Rightarrow$(1). It uses Lemma 2.2 in

D. B. A. Epstein, Curves on 2-manifolds and isotopies. Acta Math. 115 1966 83–107:

Lemma 1. Suppose that $S$ is a connected surface and $G$ is a finitely-generated subgroup of $\pi_1(S)$. Then there exists a compact $\pi_1$-injective subsurface $S'\subset S$ such that the image of $\pi_1(S')$ in $\pi_1(S)$ contains $G$.

Now, assuming that $\pi_1(S)$ is finitely-generated and $G=\pi_1(S)$, we obtain a compact subsurface $S'$ whose fundamental group is isomorphic to $\pi_1(S)$ via the map induced by the embedding $S\to S'$. Now one can either quote Goldman's result to conclude a homeomorphism $int(S')\to S$, or directly argue that each component of $S- int(S')$ is homeomorphic to $S^1\times [0,1)$:

Lemma 2. Suppose that $A$ is a connected noncompact surface with connected boundary $C=\partial A$, such that the inclusion $C\to A$ induces an isomorphism of fundamental groups. Then $A$ is homeomorphic to $S^1\times [0,1)$.

Proof. Consider the surface $B$ obtained by attaching a closed disk $D$ to $A$ along $C$. Then, by the van Kampen theorem, $B$ is simply-connected (since $\pi_1(C)\to \pi_1(A)$ is surjective). Therefore, since $B$ is noncompact, $B$ is homeomorphic to ${\mathbb R}^2$, see e.g. Corollary 1.8 in Epstein's paper. By the 2-dimensional Schoenflies theorem, there is a homeomorphism ${\mathbb R}^2\to {\mathbb R}^2$ sending $D$ to the closed unit disk $D(0,1)$ centered at the origin. This homeomorphism sends $A$ to the complement to the interior of $D(0,1)$, which is homeomorphic to $S^1\times [0,1)$. qed


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