Question

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

Answer 1

I know of several different arguments. You can decide which one you think is most elegant...

1. Rohlin's argument, which is actually quite geometric. You start with an immersion of the 3-manifold in $\mathbb{R}^5$. You modify the immersion by a cobordism until it is an embedding, and then find an explicit 4-manifold bounding it. This is nicely explained in "A la recherche de la topologie perdue". I believe this is also Richard Kent's answer above.

2. 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.

3. 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.

4. I also have a proof with Francesco Costantino, also direct and geometric. You take the compact 3-manifold 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 4-manifold. 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 3-sphere (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 4-ball.

The reason to prefer our proof (number 4) is that it is more efficient, in that (e.g.) for a 3-manifold triangulated with $n$ tetrahedra, it gives a 4-manifold with bounded geometry with $O(n^2)$ simplices. By comparison, the mapping-class group arguments of (3) tend to give a 4-manifold 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 non-explicit; I don't know how to extract any bounds from it. Rohlin's proof (1) can, I believe, be shown to give a 4-manifold with $O(n^4)$ simplices, although I never worked out all the details.

Answer 2

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 non-orientable three-manifolds into five-dimensional Euclidean space. Dokl. Akad. Nauk SSSR 160 1965 549–551.

Wall, All 3-manifolds imbed in 5-space. Bull. Amer. Math. Soc. 71 1965 564–567. )

Answer 3

MR0809959 (87f:57016) Rourke, Colin . A new proof that $\Omega_3$ is zero. J. London Math. Soc. (2) 31 (1985), no. 2, 373--376.

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 3-manifold 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 3-manifold $M(\mathbf{x},\mathbf{y})$ is obtained from $S(\mathbf{x},\mathbf{y})$ by attaching thickened 2-discs 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 3-balls. The existence of a Heegaard diagram for any 3-manifold 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 3-manifold bounds a 4-manifold 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.