Restrictions on homology for compact orientable 3-manifolds I am working on the following problem in May's Concise:

Let $M$ be a compact connected orientable $3$-manifold with boundary $\partial M \neq \emptyset$. If $\partial M$ contains no $2$-spheres, then $H_1(M; \mathbb{Z})$ is infinite. (p.172)

I am slightly confused since the following argument seems to prove an even stronger result that if $H_1(M; \mathbb{Z})$ is finite, then $\partial M$ must only consist of $2$-spheres:

If $H_1(M; \mathbb{Z})$ is finite, by universal coefficients $H_1(M; \mathbb{Q}) = 0$. Poincaré duality gives $H^1(M; \mathbb{Q}) \cong H_2(M, \partial M; \mathbb{Q}) = 0$. Considering the long exact sequence of the pair $(M, \partial M)$, this implies $H_1(\partial M; \mathbb{Q}) = 0$. By the classification theorem for closed orientable $2$-manifolds, $\partial M$ must be a disjoint union of $2$-spheres.

Does this argument fail at some point? This is essentially how I proved the previous problem (which is to show that $\partial M$ is a disjoint union of spheres given $H_1(M; \mathbb{Z}) = 0$) and I am expecting this particular problem to have a different content.
 A: Your proof is correct.
The result you proved is also a consequence of an important relationship between a 3-manifold and its boundary, which might be worth mentioning.
Proposition. Let $M$ be a compact orientable $(4n+3)$-manifold and denote by $i : \partial M \to M$ the inclusion. Then $K = \ker\left(i_* : H_{2n+1}(\partial M) \to H_{2n+1}(M)\right)$ is a Lagrangian subspace of $(H_{2n+1}(\partial M), \omega)$, where $\omega : H_{2n+1}(\partial M) \otimes H_{2n+1}(\partial M) \to \mathbb Z$ is the intersection product.
(If you have a free f.g. Abelian group $A$ with a regular skew symmetic form $\omega$, a Lagrangian subspace is a submodule $L \subset M$ such that $L = \left\{ x \in A \, \middle|\, \forall y \in L, \omega(x,y) = 0\right\}$; it's always a free f.g. Abelian group of rank $\frac 12 \mathop{\mathrm{rk}}A$.)
This result is famous, and it's a consequence of Poincaré duality not unlike your own proof. I haven't been able to find a good, simple, reference, though...
How does that prove your result? Well, if $\partial M$ is not a bunch of 
$2$-spheres, the remark on the rank of $K$ proves that $H_{2n+1}(\partial M)/K$ is infinite, which concludes, since this quotient embeds in $H_{2n+1}(M)$.
In fact, there is no need to prove the whole above proposition. If you prove that $K$ is isotropic, which means that $\forall x, y \in K, \omega(x,y)=0$, that already implies that $K$ is a free f.g. Abelian group of rank at most $\frac 12 \mathop{\mathrm{rk}} H_{2n+1}(M)$, and the end of the proof works perfectly.
At least in dimension $3$ ($n = 0$), there is a very nice topological viewpoint on this isotropy: if you take two simple curves $C$, $C'$ on $\partial M$ whose homology classes are in $K$, then $C$ bounds a properly embedded orientable surface $S \subset M$ (and ditto for $C'$, which bounds $S'$). Up to jiggling things a bit, you may assume that all intersections are transverse. In particular $S \cap S'$ is a compact $1$-manifold. Some of its components are circles, and we don't care about them. The others are arcs going from one point in $C \cap C'$ to another, and every point in $C \cap C'$ is thus paired with another one. It is then quite easy to note that two such paired points contribute to the intersection number $\omega([C],[C'])$ with opposite signs (one contributes $1$, the other $-1$), so that $\omega([C],[C']) = 0$.
