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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?

Gleb
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  • $H_(X, A) = \tilde{H}_(cof(A \to X))$ 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
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    https://mathoverflow.net/questions/215018/representing-classes-in-relative-homology-by-submanifolds – 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
  • Yes, I realized after posting my answer that the relative case does follow from the theorem. In more detail: Let $X//A$ be $X\cup_ACA$, where $CA$ is the cone. An element of $H_n(X,A)$ gives an element of $H_n(X//A,CA)$. Choose an element of $H_n(X//A)$ mapping to it. By the theorem, $k$ times this is represented by a map $M\to X//A$ from a closed oriented $n$-manifold. $M$ may be taken to be $M'\cup_{\partial } M''$, union of two manifolds along their common boundary, with $M'$ mapping into $X$ and $M''$ mapping into $CA$. Then the homology class is represented by $(M',\partial)\to (X,A)$. – Tom Goodwillie Mar 01 '18 at 02:50
  • ($k$ times the original class, I mean) – Tom Goodwillie Mar 01 '18 at 02:57

1 Answers1

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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 $E^\infty$

Tom Goodwillie
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  • 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 01 '18 at 02:54