5
$\begingroup$

Thom's theorem states that for every homology class $\alpha \in H_{*}(X)$ there exists an integer $k = k(\alpha)$ such that the class $k\, \alpha$ comes from the fundamental class of an orientable closed smooth manifold, where $X$ is an arbitrary topological space. To be on the safe side we assume that $X$ is a countable CW-complex.

Given a topological pair $(X,A)$, $A$ is closed, and a homology class $\alpha \in H_{*}(X,A)$. Is it possible to realize $k\, \alpha$ by a map from an orientable manifold with boundary, for $k$ sufficiently large?

$\endgroup$
  • $\begingroup$ $H_*(X, A) = \tilde{H}_*(cof(A \to X))$ so the relative result follows from the non-relative result. $\endgroup$ – Dylan Wilson Feb 28 '18 at 16:11
  • $\begingroup$ I'm sorry, what is that you denote by $cof$? $\endgroup$ – Gleb Feb 28 '18 at 16:51
  • $\begingroup$ @Gleb The mapping cone of $A\to X$ (also known as the cofiber). I should add that I do not see immediately how to go from the non-relative to the relative case. $\endgroup$ – Denis Nardin Feb 28 '18 at 17:02
  • 1
    $\begingroup$ mathoverflow.net/questions/215018/… $\endgroup$ – Chris Gerig Feb 28 '18 at 17:44
  • $\begingroup$ @DenisNardin yeah you're right- the non-relative case would tell you that you can represent elements in homology by maps from manifolds to X/A, but then you have to do something like in Chris's comment to fix that to a map from a pair of manifolds or something. In any case, Tom's answer is definitive... I was just trying to find one that's more elementary $\endgroup$ – Dylan Wilson Feb 28 '18 at 19:02
9
$\begingroup$

Yes. It's quite general. If $h$ is any generalized homology theory (for example, oriented bordism) such that $h_0(point)=\mathbb Z$ and such that $h_n(point)=0$ for all $n<0$ then there is a map from $h$ to ordinary homology inducing an isomorphism $h_0\to H_0$; and after tensoring with $\mathbb Q$ this map $h\to H$ becomes surjective for all spaces and pairs. You can see this using the Atiyah-Hirzebruch spectral sequence with $E^2_{p,q}=H_p(X,A;h_q(point)\otimes \mathbb Q)$, abutting to $h_{p+q}(X,A)\otimes \mathbb Q$. All of the differentials are zero because they are essentially stable operations on rational homology, so the edge $E^2_{p,0}$ survives to $E^\infty$

$\endgroup$
  • $\begingroup$ Thank you very much for your answer. Unfortunately, I am not that familiar with generalized homology theories. I'll definitely try learn more about them. Meanwhile... Could you please be so kind as to expand you answer a bit? add some comments? $\endgroup$ – Gleb Feb 28 '18 at 16:00
  • $\begingroup$ The relative version does follow from the absolute version (see comments to question). Even in the absolute case, the argument in my answer is how I think of why the theorem is true. I don't remember how Thom explained it in his original paper (before generalized homology and Atiyah-Hirzebruch SS). $\endgroup$ – Tom Goodwillie Mar 1 '18 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.