What is a higher genus analogue of the Pontryagin product? Given a compact oriented aspherical $3$--manifold $M$ with torus boundary $\partial M\simeq T^2$ (e.g. a knot complement), the condition that the images in $\pi_1 M$ of basis $x,y\in \pi_1 T^2$ under the inclusion generate the boundary of a compact oriented $3$-manifold with fundamental group $\pi_1 M$ (the $3$-manifold $M$ itself, to be precise) is algebraically encoded by the vanishing of their Pontryagin product $\langle x,y\rangle\in H_2 (\pi_1 M)$, which you can think of as the fundamental class of the torus under the inclusion. Recall that the Pontryagin product sends two commuting elements of a group to an element of $H_2$ of the group.

Question: Given a compact oriented aspherical $3$--manifold $M$ with boundary a closed surface $\partial M \simeq \Sigma$ of genus $g\geq 2$, how can I express the statement that $\Sigma$ bounds a $3$--manifold (the $3$-manifold $M$) on the level of fundamental groups? What condition must images under the inclusion of a basis for $\pi_1 \Sigma$ satisfy? How can I express the fundamental class of $\Sigma$ in terms of elements of $\pi_1$, if I know what $\Sigma$ is topologically?

I'm trying to identify some homomorphs of the fundamental group of a knotted theta, with peripheral data, and the answer to this question would, I presume, be something all such homomorphs would have to satisfy.
 A: An obvious necessary condition is that the pair $(M,\Sigma)$ satisfies Lefschetz duality. Since $M$ and $\Sigma$ are aspherical, this can be reformulated in terms of group cohomology.
Let $\Gamma=\pi_1(M)$ and $S=\pi_1(\Sigma)$. Assume that the inclusion $\Sigma\subset M$ is $\pi_1$-injective, so that $S$ is a subgroup of $\Gamma$. Then the pair $(\Gamma,S)$ is a Poincaré duality pair of dimension $3$. See Definition 4.6 of this survey by Davis, which refers to papers of Bieri and Eckmann. 
This should place some pretty serious restrictions on how the homology of $\Sigma$ sits in the homology of $M$, from which you might be able to extract a useful answer to your question. 
A: Is there such a statement?   I don't think so. As long as your manifold supports homotopy equivalences that do not restrict to homotopy equivalences on the boundary you have this problem.  So manifolds with compressible boundary are a problem like high genus handelbodies - the handelbody mapping class group is very different from the outer automorphism group of a free group. So there is no such result for these manifolds.  You'll want a more restrictive class of three manifolds to get a statement.
