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Let $X$ be a topological space (say a manifold). A result of R. Thom states that the pushforwards of fundamental classes of closed, smooth manifolds generate the rational homology of $X$. This work of Thom predates the development of bordism. Is there now a more elementary proof of this result that does not rely on spectral sequence techniques?

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"This work of Thom predates the development of bordism". That's strange thing to say, as the relevant paper of Thom ("Quelques proprietes...") is the one that introduces (co)bordism theory - and takes several giant strides in developing it. [Perhaps you're referring to the Atiyah-Hirzebruch s.s., which did come later?] –  Tim Perutz Oct 25 '11 at 16:21
    
I am confused by the question. Surely Thom's proof used bordism. Maybe by "the development of bordism" you mean the development of generalized homology theories using the idea of bordism? –  Tom Goodwillie Oct 25 '11 at 16:28
    
"you mean the development of generalized homology theories using the idea of bordism? " Yes, I mean exactly that. I was once told by an algebraic topologist that once bordism was developed as a generalized homology theory the proof of Thom's result is "obvious". However, I have so far failed to construct this "obvious" argument. Perhaps it is well known? –  Lost Oct 25 '11 at 16:40
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Thom doesn't use spectral sequences in his original paper... –  Dylan Wilson Oct 25 '11 at 18:47

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A nice, direct combinatorial construction was given by Gaifullin, see his papers on the arXiv. A drawback of this approach is that if you think of it as realizing some multiple $m\alpha$ of the given integral homology class $\alpha$ by an oriented smooth manifold, then $m$ is not bounded in terms of the dimension of $\alpha$.

There has also been another geometric approach. Thom also proved that $\bmod2$ homology classes are representable by maps of smooth (possibly unorientable) manifolds. This was reproved geometrically in

S. Buoncristiano and D. Hacon, An elementary geometric proof of two theorems of Thom, Topology 20 (1981), no. 1, 97–99

The other theorem of their title is that unoriented bordism is determined by Stiefel-Whitney numbers, and it is used in their proof that mod 2 homology classes are representable by smooth manifolds.

I believe the same geometric argument should also work to show that rational homology classes are representable by oriented smooth manifolds - modulo the fact that Pontryagin numbers determine oriented bordism tensored by $\Bbb Q$. This fact I'm afraid I don't know how to prove geometrically (for some proof, see e.g. the Milnor-Stasheff book). But note that in a subsequent paper Buouncristiano and Hacon also gave a geometric proof that Chern numbers determine complex bordism (Ann. of Math., 118 (1983), 1-7). Their other papers may also be of interest if you care about geometric proofs of classical results on bordism.

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I suspect that the "obvious" proof used an Atiyah-Hirzebruch spectral sequence, so it's not obvious unless you are happy with spectral sequences.

Here is an argument with no spectral sequence in it.

There is a homology theory $\Pi_\ast$ called stable homotopy theory. It has a natural map to ordinary homology $H_\ast$, given by the Hurewicz map. After tensoring with $\mathbb Q$ this map $\Pi_\ast(X)\to H_\ast(X)$ becomes an isomorphism. The proof of this for finite complexes $X$ uses the five lemma plus the fact that it is an isomorphism when $X$ is a point. In the case when $X$ is a point this is the result (Serre's thesis) that $\pi_k(S^n)$ is finite if $k>n$.

On the other hand, by the Thom-Pontryagin construction, stable homotopy is the same as framed bordism.

(Oh wait, Serre's result used a spectral sequence ...)

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I think that Thom's original argument was also based on the computation of homotopy groups of spheres so the argument you sketch is probably not all that different under the surface. Does the proof based on the Atiyah-Hirzebruch spectral sequence use this computation? –  Lost Oct 25 '11 at 17:45
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I suppose it must. In the background of all of this stuff, in effect, is the fact that homology groups of spectra are rationally the same as homotopy groups of spectra. I chose to address the rational isomorphism from framed bordism to homology rather than the rational surjection from oriented bordism to homology. This means I could use the five lemma. It also means that I could begin with that fact for the sphere spectrum and deduce it for suspension spectra, but I never needed Thom spectra. –  Tom Goodwillie Oct 25 '11 at 18:37
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As an educational remark, I would like to mention the recent book by Tammo Tom Dieck {Algebraic Topology (EMS Textbooks in Mathematics)} that proves many of the standard theorems of algebraic topology (including Serre's computation of rational homotopy groups) without the use of spectral sequences. I suppose in the question above I am interested in a direct geometric argument that does not secretly involve homotopy groups of spheres. –  Lost Oct 25 '11 at 20:00
    
Just a small correction, when you say that $\pi_k(S^n)$ is finite if $k>n$, you should exclude the case of $\pi_{4k-1}(S^{2k})$ which is $\mathbb{Z}$ plus finite. –  YangMills Nov 2 '11 at 18:13
    
Yes, of course. Maybe I should have said if $n<k<2n-1$. –  Tom Goodwillie Nov 2 '11 at 21:42

Thank you everyone for helping with this question. I would like to attempt to provide my own answer (which came to me after reading all the comments): Let $B_* (M)$ be rational, oriented bordism and $H_* (M)$ be the rational homology of $M$. I claim there is a map $$F:B_* (M) \rightarrow H_* (M)\otimes B_* (pt)$$ that is an isomorphism. The map $F$ sends $(P \rightarrow M) $ to $([P ] \otimes 1+1\otimes P')$ where $[P]\in H_* (M) $ represents the fundamental class of $P$ and $P' \in B_* (pt)$ is the bordism element represented by the abstract manifold $P$. This map is clearly an isomorphism when $M= pt$. The standard inductive argument over the number of cells implies this is an isomorphism in general. Note that no appeal to homotopy groups, spectral sequences or Thom spectra is being implictly used. We do use the computation of 0 dimensional bordism groups.

Is this correct?

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No, this (the argument, not the conclusion) is incorrect. Because the complex cobordism ring is torsion-free, the argument, if correct, would apply to (integral) complex cobordism as well (the target of $F$ is a homology theory). It would follow that $MU_{\ast} (M)= H_{\ast}(M;MU_{\ast})$. In other words, the complex cobordism spectrum would be a wedge of Eilenberg-MacLane spectra, which is wrong. The mistake in the argument is that the proposed $F$ is not a transformation of homology theories (it does not commute with the boundary maps). –  Johannes Ebert Nov 2 '11 at 20:02

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