TITLE: Torsion in cuspidal cohomology
QUESTION [10 upvotes]: Following Lemma 2.7 from Vogtmann's Rational Homology of Bianchi Groups, I want to define cuspidal cohomology as $$H_{\mathrm{cusp}}(M)=\frac{H_1(M)}{i_*(H_1(\partial M))}$$ where $i:\partial M\to M$ is the inclusion and I take integral coefficients. The motivation is that Abelian covering spaces such that the preimage of a cusp consists of disjoint copies of the same cusp are corresponding to exactly those holonomies $\pi_1(M)\to G$ that factor through the cuspidal cohomology.
I verified for all hyperbolic cusped 3-manifolds in the SnapPea census that if $H_1(M)=\mathbb{Z}^n\oplus T$ where $T$ is torsion, then $H_\mathrm{cusp}(M)=\mathbb{Z}^{n-\mbox{number of cusps}}\oplus T$. This is to be expected rationally by half-lives-half-dies. It seems to also hold integrally for hyperbolic 3-manifolds.
Is there an argument that this holds for all hyperbolic cusped 3-manifolds? (And maybe a counterexample if you take a non-hyperbolic manifold)

REPLY [10 votes]: Consider the long exact sequence on homology (coefficients in $\mathbb{Z}$)
$$\to H_1(\partial M)\overset{i}{\to} H_1(M)\to H_1(M,\partial M) \to H_0(\partial M) \to $$
You are looking for $H_1(M)/i_\ast(H_1(\partial M)) \cong im\{ H_1(M)\to H_1(M,\partial M)\} \cong ker \{ H_1(M,\partial M)\to H_0(\partial M)\}$ by the exactness of the sequence. 
If $H_1(M)\cong \mathbb{Z}^n\oplus  T$, $T$ torsion, then $H_2(M)\cong \mathbb{Z}^{n-1}$ (the rank $n-1$ follows from Euler characteristic). By universal coefficients, we have a short exact sequence
$$0\to Ext(H_1(M),\mathbb{Z})\to H^2(M) \to Hom(H_2(M),\mathbb{Z})\to 0.$$
One computes $Ext(H_1(M),\mathbb{Z})\cong T$, $Hom(H_2(M),\mathbb{Z})\cong \mathbb{Z}^{n-1}$ (see p. 195 of Hatcher), so that $H^2(M)\cong \mathbb{Z}^{n-1} \oplus T$. By Lefschetz duality, $H^2(M)\cong H_1(M,\partial M) \cong \mathbb{Z}^{n-1}\oplus T$. So you are looking for the torsion in 
$$ker\{ H_1(M,\partial M)\to H_0(\partial M) \} \cong ker\{ \mathbb{Z}^{n-1}\oplus T \to \mathbb{Z}^c\},$$
which is clearly isomorphic to $T$.