I'm looking for an elegant proof that any closed, oriented $3$manifold $M$ is the boundary of some oriented $4$manifold $B$.

3$\begingroup$ Kirby's book "Topology of 4manifolds" or Gompf and Stipsicz's book "4manifolds and Kirby Calculus" both have fairly elementary proofs of this. I believe one of the earliest proofs is in Thom's big paper on cobordism: R. Thom, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28, 1786 (1954). $\endgroup$ – Ryan Budney Apr 29 '11 at 3:20

1$\begingroup$ Depending on what you mean by elegant there are also some more modern proofs that are closer to algorithmic questions, like Costantino and D.Thurston's work. $\endgroup$ – Ryan Budney Apr 29 '11 at 3:22

1$\begingroup$ A link to Costantino and Thurston's article: arxiv.org/abs/math/0506577 $\endgroup$ – j.c. Apr 29 '11 at 13:27

$\begingroup$ @Ryan: Gompf and Stipsicz give brief descriptions of Rohlin's proof and Lickorish's proof, both described in the answers below. I haven't had a chance to look at Kirby's book yet. $\endgroup$ – Dylan Thurston May 6 '11 at 9:54

1$\begingroup$ See a more general discussion here mathoverflow.net/q/306506/27004 $\endgroup$ – wonderich Oct 30 '18 at 20:43
I know of several different arguments. You can decide which one you think is most elegant...
Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Autumn Kent's answer above.
Thom's argument, with lots of algebraic topology. This is probably not the most elegant route if you only want this piece, although of course Thom tells you much more.
Rourke's argument as sketched by Daniel Moskovich above. Indeed, any proof that the mapping class group is generated by Dehn twists also gives a proof that $\Omega_3 = 0$. Dehn and Lickorish also have proofs of this.
I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3manifold and look at a generic map to $\mathbb{R}^2$. The preimage of a generic point is a disjoint union of circles, which bounds a convenient canonical surface (a union of disks). Take these disks as the start of your 4manifold. In codimension one singularities, two of these circles can merge, and the preimage of a little transversal is a pair of pants, which can be filled in with a 3sphere (together with the disks already attached). In codimension 2, there are only two different interesting local models, and both can be filled in canonically with a 4ball.
The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3manifold triangulated with $n$ tetrahedra, it gives a 4manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mappingclass group arguments of (3) tend to give a 4manifold of complexity at least exponential in $n$, and usually a tower of exponentials. (You can see this already in the inductive argument sketched out in Daniel Moskovich's answer.) Thom's proof (2) is completely nonexplicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4manifold with $O(n^4)$ simplices, although I never worked out all the details.

$\begingroup$ Can you apply the argument 4 to a fourmanifold $M^4$? Namely, map $M^4$ to $\mathbb R^3$ and apply the same method to construct a 5manifold bounded by $M$. Of course, at some point you should need to do some topological blowups on $M^4$ to get zerosignature. $\endgroup$ – Bruno Martelli Apr 30 '11 at 9:28

1$\begingroup$ @Bruno: Yes, you can. You end up showing any 4manifold is cobordant to a connect sum of $\mathbb{CP}^2$'s and $\overline{\mathbb{CP}}^2$'s. $\endgroup$ – Dylan Thurston Apr 30 '11 at 9:43

1$\begingroup$ I confess that I never understood the ideas of Thom's proof, certainly not well enough to see that it is fundamentally nonexplicit. $\endgroup$ – Greg Kuperberg May 1 '11 at 17:12

$\begingroup$ @Greg: I only understood Thom's proof well enough to see that it would be at least quite difficult to make it explicit, at least difficult enough that an explicit version might well be called a different proof. $\endgroup$ – Dylan Thurston May 2 '11 at 3:39

2$\begingroup$ Can you use your argument to bound the number of components of a link needed in a Kirby diagram? This gives some estimate on the minimal rank of H_2 of a simplyconnected 4manifold bounding the 3manifold. $\endgroup$ – Ian Agol May 5 '11 at 21:56
Thom wrote two notes in the proceedings of the "Colloque de Topologie de Strasbourg", which was a topology seminar organized by Ehresmann at that time:
"Quelques propriétés des variétésbords", Colloque de Topologie de Strasbourg, 1951, no. V. La Bibliothèque Nationale et Universitaire de Strasbourg, 1952.
"Sur les variétés cobordantes", Colloque de topologie et géométrie différentielle, Strasbourg, 1952, no. 7. La Bibliothèque Nationale et Universitaire de Strasbourg, 1953.
These two notes prefigure some of the results and some of the techniques that will appear slightly later in his 1954 paper "Quelques propriétés globales des variétés différentiables" and each of these notes contains a proof of the fact that $\Omega_3=0$. In fact, these two prior proofs are different from the demonstration that follows by specializing the results of his '54 paper, and which was alluded to in the above answers.
His '52 proof (in the 1953 proceedings) is close to Pontryagin's ideas relating framed cobordism groups to stable homotopy groups of spheres. Thom considers there the map $\psi_k:\pi_{n+k}(S^n) \to \Omega_k$ which consists in taking regular preimages of smooth approximations of maps $S^{n+k} \to S^n$. This map is additive and its image consists of the cobordism classes of those $k$manifolds which can be embedded in $\mathbb{R}^{n+k}$ with trivial normal bundle. Since any closed orientable $3$manifold is parallelizable, the map $\psi_3$ is surjective (for $n$ large enough) and it suffices to prove that $\psi_3$ vanishes on a generating set of the group $\pi_{n+3}(S^n)$. To conclude his argument, Thom mentions that this stable homotopy group is isomorphic to $\mathbb{Z}_{24}$ (which, I guess, he knew from Cartan or Serre) and, most importantly, that it is generated by the suspension of the quaternionic Hopf fibration with fiber $S^3= \partial D^4$.
In contrast with his '52 and '54 proofs, his '51 proof (in the 1952 proceedings) does not use any kind of PontryaginThom constructions but, as Lickorish and Fenn will do some years later, he uses the fact that any closed oriented $3$manifold can be presented by a Heegaard splitting. Let $H_g$ be the handlebody of genus $g$, let $\hbox{MCG}(\partial H_g)$ denote the mapping class group of its boundary and for any $f \in \hbox{MCG}(\partial H_g)$, set
$$ V_f := H_g \cup_f (H_g). $$
Thom starts by observing that, if $V_f$ and $V_{f'}$ bound, then so does $V_{f \circ f'}$ (see Bruno's comment to Daniel's answer). Next, he recalls that Dehn found an explicit and finite system of generators $\{\tau_i\}$ for $\hbox{MCG}(\partial H_g)$ which consists of several Dehn twists: there are 2, 5 and 2g(g1) such generators for $g=1$, $g=2$ and $g>2$ respectively. Thus, by the previous observation, it suffices to check that $V_{\tau_i}$ is a boundary for any $g\geq 1$ and for any $i$. For any $g>1$, it can be observed that each of the Dehn twists $\tau_i$ is performed along a curve which avoids a full (solid) handle of $H_g$: hence $V_{\tau_i}$ is the connected sum of $S^1 \times S^2$ with a $3$manifold admitting a Heegaard splitting of genus $g1$. Since such a connected sum can be realized in dimension $4$ by the attachement of a handle of index $1$, we are allowed to do an induction on the genus $g$. For $g=1$, the $3$manifolds $V_{\tau_1}$ and $V_{\tau_2}$ are $S^3$ and $S^1 \times S^2$, which are obviously boundaries.
It seems that this early proof that $\Omega_3=0$ has been forgotten since then: this is probably due to the fact that the proceedings is not widely available and is written in french. Personally, I did not know it before I opened this proceedings in the Math' Library of Strasbourg a few months ago! Nonetheless, Haefliger mentions it in his survey paper of Thom's works (Publ. IHES 1988).
Maybe overkill, but elegant:
By a theorem of Hirsch, an oriented $3$manifold $M$ embeds in the $5$sphere (nonorientable case: Rohlin and Wall, independently). By Alexander duality, $M$ bounds a "Seifert $4$manifold."
(Some references:
Hirsch, Immersions of almost parallelizable manifolds. Proc. Amer. Math. Soc. 12 1961 845–846.
Rohlin, The embedding of nonorientable threemanifolds into fivedimensional Euclidean space. Dokl. Akad. Nauk SSSR 160 1965 549–551.
Wall, All 3manifolds imbed in 5space. Bull. Amer. Math. Soc. 71 1965 564–567. )

5$\begingroup$ It's funny, I sorta thought this was how everybody thought about it. I guess having Cameron Gordon as an advisor colors your world view. $\endgroup$ – Autumn Kent Apr 29 '11 at 13:52

10$\begingroup$ You can avoid Hirsch's theorem by using Whitney's immersion theorem of $M^n$ in $\mathbb R^{2n1}$, which implies that every 3manifold $M$ immerses in $\mathbb R^5$. Such an immersion can be perturbed so that selfintersections have dimension 3+35 = 1, i.e. are circles. You can then surger the manifold around the circles to get an embedding, and such a surgery can easily be realized by a cobordism. Then you use Alexander duality. This was Rohlin's original argument. $\endgroup$ – Bruno Martelli Apr 29 '11 at 16:34

2$\begingroup$ This is beautifully intuitive! But it's definitely massive overkill each step is in itself more difficult that the fact being proven. Statement (1) is the easiest statement among: 1) Omega_3=0 2)Whitney immersion theorem 3)Hirsch's theorem 4) Alexander duality in this context. This is highlighted by Rourke's short elementary proof (which isn't half as conceptually satisfying for me, though). $\endgroup$ – Daniel Moskovich Apr 29 '11 at 16:47

2$\begingroup$ @Greg: It is related, though in a different direction; see my answer. @Daniel: I disagree that Whitney immersion is hard, and Hirsch's theorem is not necessary, as Bruno sketches. $\endgroup$ – Dylan Thurston Apr 29 '11 at 23:37

2$\begingroup$ You can also use an embedding in S^6 (a bit easier to construct). If nM = normal bundle, then H^2(S^6nM) = Z by Mayer Vietoris. A geometric representative of the generating class will be a 4manifold X with dX = M. More precisely, dX will be cobordant to a nonvanishing section of nM. $\endgroup$ – Lucas Culler Dec 6 '11 at 3:30
MR0809959 (87f:57016) Rourke, Colin . A new proof that $\Omega_3$ is zero. J. London Math. Soc. (2) 31 (1985), no. 2, 373376.
Edit: To summarize: Rourke's proof is short and elementary. Other proofs which I know involve either significant algebraic topology which is much harder than the theorem (Thom, or Rohlin), or lengthy calculations (Lickorish).
Any orientable 3manifold has a Heegaard diagram $S(\mathbf{x},\mathbf{y})$, where $S$ is an orientable surface with two complete systems of curves $\mathbf{x}$ and $\mathbf{y}$ (a system of curves is complete if each curve it contains is simple and closed, its curves are pairwise disjoint, and their union does not separate $S$). A closed orientable 3manifold $M(\mathbf{x},\mathbf{y})$ is obtained from $S(\mathbf{x},\mathbf{y})$ by attaching thickened 2discs along $\mathbf{x}$ on $S\times {\{0\}}$ and along $\mathbf{y}$ on $S\times {\{1\}}$, and then filling in the resulting $S^2$ boundaries with 3balls. The existence of a Heegaard diagram for any 3manifold is a reformulation of the elementary fact that it has a handle decomposition.
Rourke's proof is by induction on the genus $g$ of $S$ and on the minimum intersection number $r$ of a curve in $\mathbf{x}$ with a curve in $\mathbf{y}$. Namely, he gives a straightforward combinatorial argument for why, if $r>1$, then there exists a third complete system of curves $\mathbf{z}$ on $S$ whose minimum pairwise linking with both $\mathbf{x}$ and $\mathbf{y}$ is less than $r$. Surgery of $M(\mathbf{x},\mathbf{y})$ around $\mathbf{z}$ gives the connect sum of $M(\mathbf{x},\mathbf{z})$ and $M(\mathbf{z},\mathbf{y})$. Finally, if $r\leq 2$, then you can chop one off the genus of $S$ pretty easily. And that's all there is to it, by induction.
Rourke's proof makes the fact that any closed orientable 3manifold bounds a 4manifold look like a stupidly easy combinatorics exercise. It's certainly my favourite proof of this theorem, although other ways of looking at the problem are not without their charm.

2$\begingroup$ Nice induction proof. You can think of 3 handlebodies attached to the same surface $S$: if two pairs of them bound 4manifolds $M_1, M_2$, then also the third pair does (the union of $M_1$ and $M_2$ along the third handlebody). $\endgroup$ – Bruno Martelli Apr 30 '11 at 9:26