# Relative Steenrod's problem

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?

• $H_*(X, A) = \tilde{H}_*(cof(A \to X))$ so the relative result follows from the non-relative result. – Dylan Wilson Feb 28 '18 at 16:11
• I'm sorry, what is that you denote by $cof$? – Gleb Feb 28 '18 at 16:51
• @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. – Denis Nardin Feb 28 '18 at 17:02
• mathoverflow.net/questions/215018/… – Chris Gerig Feb 28 '18 at 17:44
• @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 – Dylan Wilson Feb 28 '18 at 19:02

## 1 Answer

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 $E^\infty$

• 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? – Gleb Feb 28 '18 at 16:00
• 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). – Tom Goodwillie Mar 1 '18 at 2:54