Representing rational homology by manifolds

Question

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?

Answer 1

A nice, direct combinatorial construction was given by Gaifullin, see his papers on the arXiv (equivalently: Explicit construction of manifolds realizing the prescribed homology classes, Realisation of cycles by aspherical manifolds and Configuration spaces, bistellar moves, and combinatorial formulae for the first Pontryagin class). 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 (Core pdf)

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 https://doi.org/10.2307/2006950). Their other papers may also be of interest if you care about geometric proofs of classical results on bordism.

Answer 2

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 ...)

Answer 3

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 implicitly used. We do use the computation of 0 dimensional bordism groups.

Is this correct?