Explanation for the Thom-Pontryagin construction (and its generalisations)

Question

In 1950, Pontryagin showed that the n-th framed cobordism group of smooth manifolds was equal to n-th stable homotopy group of spheres:

$$ \lim_{k \to \infty} \pi_{n+k}(S^k) \cong \Omega_n^{\text{framed}}.$$

Later on, in his 1954 paper, Thom generalises this with the now called Thom spaces, and shows that there is a similar correspondence for more general types of cobordism: manifolds with a $(B,f)$ structure on their normal bundle; for example, unoriented cobordism for $B = BO$, oriented cobordism for $B = BSO$, complex cobordism for $B = BU$, framed cobordism for $B = BI$ for the identity $I$ in $O$, etc. (Thom considers the cases $BO$ and $BSO$.)
The generalisation he arrived to, now called the Thom-Pontryagin construction, is the following:

$$\lim_{k \to \infty}\pi_{n+k}(TB_k) \cong \Omega_n^{(B,f)}, $$

where $TB$ is the Thom space of the universal bundle over $B$ given by the classifying map $B \to BO$; $TB$ is obtained by adding a point at infinity to each fiber and gluing all these added points to a single point in the total space of the bundle.

In fact, the result can be generalised further by considering cobordism as a homology theory, and one arrives at the following:

$$\Omega_n^{(B,f)}(X,Y) \cong \lim_{k \to \infty} \pi_{n+k}(X/Y \wedge TB_k),$$

where, if $Y$ is empty, $X/Y$ is the disjoint union of $X$ with a point (and $\wedge$ is the smash product). Here $\Omega_n(X,\emptyset)$ is to be understood as a relative cobordism over $X$. This clearly generalises the previous result by taking $X$ to be a point (and $Y$ empty).

Now, my question is, how do you understand the Thom-Pontryagin construction? I've seen a few mentions of a particularly visual way of understanding it, but without much to actually back this up (besides from, I remember, a few mentions of blobs of ink). The standard proofs (in Stong's Notes on Cobordism Theory or in Thom's original paper for example) are quite long and I have trouble keeping hold of my geometric intuition throughout.

Answer 1

Let $M$ be a smooth, compact $n$-manifold with a stable framing. Then Pontryagin's map works like this:

1. There is only one embedding of $M$ into $\mathbb{R}^{k+n}$ for $k \gg n$.
2. A stable framing of $M$ induces a normal framing of $M$, because the total bundle of $M$ in this embedding is trivial.
3. Make a tubular neighborhood $N$ around $M$ which is diffeomorphic to the normal bundle of $M$.
4. Make a map from $\mathbb{R}^{k+n}$ to $S^k = \mathbb{R}^k \cup \{\infty\}$ which sends the complement of $N$ to $\infty$, and sends each normal fiber to $\mathbb{R}^k$ using the framing of that fiber. The map extends to $S^{k+n}$ by sending $\infty$ to $\infty$.

You can show that a cobordism class goes to a homotopy class with basically the same construction in the next dimension. So you get a well-defined homomorphism. It takes more work to show that it's an isomorphism, but I think that it can be done with similar geometric considerations. For instance, to show that it is surjective, consider a generic smooth map from $S^{k+n}$ to $S^k$. The inverse image of $0 \in \mathbb{R}^k \subset S^k$ is a manifold, a small neighborhood of $0$ pulls back to a framed normal bundle $N$, and (roughly speaking) everything outside of $N$ can be crammed into $\infty$ with a homotopy.

Answer 2

This is an interesting question, and Greg's answer is excellent. Thinking about this was very nice!
Regarding the Pontryagin-Thom construction, as opposed to the Pontryagin construction which other comments and answers have addressed, I think Thom's original approach, whose goal was to solve Steenrod's realization problem, was quite geometric if you think of it the right way (see this question and this question, where people ask Steenrod's realization problem as an MO question).
Going one way, our initial data is a bordism class, i.e. a closed manifold M mapping into a polyhedron X, modulo cobordism. Steenrod's realization problem, by the way, is to determine which homology classes in X are realized this way for some M. Well, what Thom does first is to embed X in some $\mathbb{R}^n$ for n large enough, and take a regular neighbourhood N of X, which, as Greg points out, is diffeomorphic to the normal bundle of X. You now collapse the boundary of N to a point, and the quotient $N/\partial N$ to the Thom space of the regular neighbourhood (normal bundle) of the image of M in X. This is a map into a Thom space, so you can compose with a map into the universal Thom space. As I understand it, the point of this whole exercise is to leverage the nice properties of Euclidean space- namely the existence of the universal Thom space. So taking stock, starting with a bordism class, I have given you a map from $N/\partial N$ to the universal Thom space, whose image is a homotopy class of the right dimension. This is totally geometric- you can "see" the class as the image of $N/\partial N$ in the universal Thom space, where everything outside the image of M is mapped to the basepoint. Thom proved that this is well-defined, and that it's a surjection.
To go the other way is even easier. Starting from a class in the universal Thom space, you realize it as a map g from $N/\partial N$ to the universal Thom space, and apply the Thom isomorphism. What is the Thom isomorphism? It's the intersection of the (relative) cycle in the Thom space of $N/\partial N$ corresponding to g with the zero section (algebraically, you're multiplying by the Thom class), which gives you a closed manifold $M^\prime$ in N. Retract N to X, and you are done, modulo transversality and other technical issues.