There are nice results for representing homology classes by submanifolds, in particular for any class in $H\_i(X)$ with $i\le 6$, see [here](https://mathoverflow.net/questions/1489/cohomology-and-fundamental-classes). When $X$ is low-dimensional I can start getting explicit, but this uses Poincare duality and appeals to classifying maps and characteristic classes of bundles. **What is the right way to approach the problem for relative homology? Does the formulation of (Thom) spectra extend here?** While I do have Lefschetz duality, I don't think I can do much with bundles, and I am not comfortable with (Thom) spectra.


**Given a submanifold $A\subset X$ of a closed (smooth?) manifold, when can I (not) represent a homology class in $H\_i(X,A;\mathbb{Z})$ by some submanifold with appropriate boundary?**


Take whatever restrictions on $i$ and $\dim X$, but assume we're in a setting where the Hurewicz map isn't an isomorphism. If $i>\dim A+1$ then $H\_i(X)$ surjects onto $H\_i(X,A)$ in the LES, so we can take a submanifold which represents a homology class of $X$ to be the submanifold that represents the relative homology class.


Perhaps there is an example of an unrepresentable class in $H\_2(X,S^1;\mathbb{Z})$ with $\dim X =3$ and $S^1$ highly knotted. When $X=S^3$ this group is $\mathbb{Z}$ (as seen by the LES), and Seifert surfaces do the trick.


As a toy model, consider the equator in a sphere. Then $H\_2(S^2,S^1)\cong\mathbb{Z}^2$, as seen by reduced homology $\widetilde{H}\_2(S^2/S^1)\cong H\_2(S^2\bigvee S^2)\cong H\_2(S^2)\oplus H\_2(S^2)\cong\mathbb{Z}^2$. It looks like the classes $(1,0)$ and $(0,1)$ come from the hemispheres, and the relative Hurewicz theorem applies I think.



Representing classes in *relative* homology by submanifolds
I think the arguments are all pretty much the same. Let $X$ be a manifold, and $A$ a subset that's "tame enough" to admit a tubular / regular neighbourhood $V$, then


$$H\_i(X,A) \simeq H\_i(X,V)$$ 


by homotopy. And $H\_i(X,V) \simeq H\_i(X \setminus int(V), \partial V)$ by excision. 


But this group is isomorphic to $H^{n-i}(X \setminus int(V))$ by Poincar\'e duality (if you assume orientability). $n=dim(X)$


At this point the standard Serre-Thom representability constructions work very nicely, for example if $i=n-1$ you use that $H^1(X \setminus int(V)) \simeq [X \setminus int(V), S^1]$ and apply transversality.


