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I got fantastic answers to my previous question (about modern references for the fact that surfaces can be triangulated), so I thought I'd ask a related question. A basic fact about surface topology is that if $S$ is a noncompact connected surface, then $\pi_1(S)$ is a free group (possibly trivial or $\mathbb{Z}$). I've had a lot of people ask me for references for this fact. I know of two such references:

1) In section 44A of Ahlfors's book on Riemann surfaces, he gives a very complicated combinatorial proof of this fact.

2) This isn't a reference, but a high-powered 2-line proof. Introducing a conformal structure, the uniformization theorem shows that the universal cover of $S$ is contractible. In other words, $S$ is a $K(\pi,1)$ for $\pi=\pi_1(S)$. Next, since $S$ is a noncompact $2$-manifold, its integral homology groups vanish in dimensions greater than or equal to $2$. We conclude that $\pi_1(S)$ is a group of cohomological dimension $1$, so a deep theorem of Stallings and Swan says that $\pi_1(S)$ is free.

There should be a proof of this that you can present in a first course in topology! Does anyone know a reference for one?

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If you accept the polygonal representation of a closed surface on the plane, then a punctured surface will be represented by the same polygon with punctures easily managable to be in the interior, but now this picture clearly homotopes to a wedge of circles hence $\pi_1$ is free. – Maharana Mar 17 '10 at 6:11
Andy, one does not need uniformization to show that $S$ is Eilenberg-MacLane. Namely, the universal cover of $S$ is acyclic (it does not have homology above dimension $1$ by Poincare duality and it does not have first homology since it is simply-connected). Now Whitehead theorem gives you that the inclusion of any point into the universal cover is a homotopy equivalence. – Igor Belegradek Mar 17 '10 at 13:04
Igor - Excellent observation! Thanks! – Andy Putman Mar 17 '10 at 13:42
Maharana - I'm not talking about punctured surfaces. I'm talking about general noncompact surfaces, like a closed surface minus a Cantor set or the boundary of a regular neighborhood of an infinite graph embedded in R^3. – Andy Putman Mar 17 '10 at 13:44
ah, got it. Was thinking of a very special and easy case! – Maharana Mar 18 '10 at 3:22
up vote 30 down vote accepted

I'm reluctant to advertise, but since no one else has answered yet, I'll mention the proof on pp. 142--144 of my book Classical Topology and Combinatorial Group Theory.

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Very nice! Another nice feature is that you include a reference to a paper of Johansson from 1931 that (I guess) is the origin of this result. – Andy Putman Mar 17 '10 at 2:57
Yes, I'm confident that the result is due to Johansson. – John Stillwell Mar 17 '10 at 2:58
If I had been faster I would have proposed the same reference so you wouldn't have had to advertise your own book. I'm a fan :-) – Alon Amit Mar 17 '10 at 16:52
That book is surely in the list of books that should be advertised loud and often! :) – Mariano Suárez-Alvarez Mar 31 '12 at 0:19

If you assume the existence of smooth structure on a noncompact surface then it is easy to show the existence of a proper morse function with no local maximum.This shows that the surface is homotopic to a one dim CW complex. This is the smooth version of Igor's answer.

EDIT BY ANDY PUTMAN: Mohan isn't registered and thus isn't able to comment, but he sent me an email with more details. The result is true in all dimensions : any noncompact smooth n-manifold is homotopy equivalent to an n-1 complex. The key is to construct a strictly subharmonic morse exhaustion function. The subharmonicity prevents the function from having local maxima. Details of this can be found in his paper "Elementary Construction of Exhausting Subsolutions of Ellitpic Operators", which was joint with Napier and was published in L’Enseignement Math´ematique, t. 50 (2004), p. 1–24.

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Is there a topological version? – Mariano Suárez-Alvarez Mar 17 '10 at 15:17
Is this a general fact about noncompact manifolds, or is this special to surfaces? I haven't managed to prove it, but that doesn't mean that it isn't as easy as you claim it is... – Andy Putman Mar 17 '10 at 15:34
It would be nice to have a good source of information on proper Morse functions; I could not find any reasonable account, and I suspect there may be some subtle issues there. But I agree with Mohan that Morse theory argument should work for the problem at hand, and I think it should work in any dimension. – Igor Belegradek Mar 17 '10 at 16:05

I just ran across this question, and thought I would give a precise version of the proof Ilya suggested. I believe I learned this proof in Richie Miller's topology course, Michigan State University, 1977 or so.

Choose a triangulation of the surface $S$, equipped with the simplicial metric. Choose a maximal one-ended subtree $T$ of the dual 1-skeleton $S^{(1)}$. The subtree $T$ contains every dual $0$-cell, that is, the barycenter of every 2-simplex. Also, $T$ contains dual 1-cells crossing certain $1$-simplices. Let $U$ be the union of the open 2-simplices and open 1-simplices that contain a point of $T$. The metric completion of $U$, denoted $\bar U$, is a closed disc with one boundary point removed, and so there is a deformation retraction from $\bar U$ onto its boundary $\partial \bar U$. Attaching $\bar U$ to $S - U$ in the obvious way to form the surface $S$, the deformation retraction $\bar U \to \partial\bar U$ induces a deformation retraction of $S$ onto $S-U$, wnich is a subcomplex of the 1-skeleton.

By the way, the subtree $T \subset S^{(1)}$ can be constructed by an explicit process. Enumerate the dual $0$-cells $v_1,v_2,\ldots \in S^{(1)}$. Construct one-ended subtrees $T_1,T_2,\ldots \subset S^{(1)}$ as follows. $T_1$ is any proper ray based at $v_1$. If $v_n \in T_{n-1}$ then $T_n = T_{n-1}$. If $v_n \not\in T_{n-1}$, let $T_n$ be the union of $T_{n-1}$ with any arc $\alpha \subset S^{(1)}$ having one endpoint at $v_n$ and intersecting $T_{n-1}$ in its opposite endpoint. Each $T_n$ is a one-ended tree by induction, and since the radius $r$ neighborhood of $v_1$ in $T_n$ stabilizes as $n \to \infty$, it follows that $T = \cup_n T_n$ is a one-ended subtree of $S^{(1)}$, and it is maximal because it contains each $v_i$.

I think this proof generalizes to any dimension, to give the theorem that Igor Belegradek refers to.

--- Edited to simplify and clarify the argument ---

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This is a beautiful proof. Thanks Lee! – Andy Putman Mar 30 '12 at 23:46

Not to take anything away from the other answers, but I believe that the result (the more general n-dimensional version mentioned by @Igor Belegradek) is actually due to J. H. C. Whitehead: The immersion of an open 3-manifold in euclidean space, Proc. London Math. Soc 11 1961, 81-90., lemma 2.1 (JHC was a little modest calling this a lemma).

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@Igor: That's a nice reference! Actually, Whitehead's proof is essentially the same as the one outlined by Lee. – Misha Mar 31 '12 at 3:10
@Misha: Thanks! I did notice that this was the same argument -- I guess Whitehead found the canonical argument... – Igor Rivin Mar 31 '12 at 4:30

I don't have a reference but here is one way to prove what you want. It is a basic result of PL-topology that any open PL-manifold deformations retracts to a subcomplex of lower dimension. Thus you are reduced to showing that the fundamental group of a graph is free, so collapse a maximal subtree to a point, get a wedge of circles and apply Van Kampen.

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Do you (or anyone else) know a reference for this "basic fact in PL-topology". I just skimmed through Rourke-Sanderson and didn't find it. – Andy Putman Mar 17 '10 at 3:19
Andy, this fact was mentioned in Brown's book "Cohomology of groups" in the proof of Proposition 8.1 (Chapter 8, Section 8). Brown does not give a reference. I do not know how the proof goes, and also would be interested in a reference. – Igor Belegradek Mar 17 '10 at 12:07
@Andy: This doesn't seem to be so hard if we can assume that the surface is a compact triangulated surface with some points removed. This is the same as removing the interiors of a few triangles. Then, whenever a see an edge that bounds a triangle on only one side, we remove both the edge and the interior of that triangle. This is a deform. retraction. The only way that this process can stop is when there are no filled-in triangles (assuming the surface was connected), so we have something 1-dimensional left. Did I miss something? Or do we need some generalization of this in your case? – Ilya Grigoriev Mar 17 '10 at 17:37
@Ilya - that is the trivial case! There are a whole zoo of non-compact surfaces that are nowhere close to these examples. For instance, take a compact surface and remove a cantor set. I want arguments that handle those cases as well. – Andy Putman Mar 17 '10 at 18:06

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