$H_2(\partial V)=0$ if $V$ is a rational homology 4-ball

Question

I’m reading “On knots” by L. Kauffman. In Chapter XVII (which follow the prominent paper of Casson and Gordon,) we have the following lemma.

Lemma 17.3. Let $V$ be a $\mathbb{Q}$-homology 4-ball. If the image of $H_1(\partial V)\to H_1(V)$ has order $l$, then $H_1(\partial V)$ has order $l^2$.

In the very first part of the proof, we have the following series of isomorphisms. $$ H_2(\partial V) \cong H^1(\partial V)\cong \mathrm{Hom}(H_1(\partial V),\mathbb{Z})=0$$

Okay, the first isomorphism is the Poincaré duality for the closed manifold $\partial V$, and the second isomorphism is the universal coefficient theorem. The third isomorphism is the one I cannot see. It seems to be relevant with the fact that $V$ is a rational homology 4-ball, but how should I extract information on the homology of $\partial V$ from this? Thank you in advance.

Answer 1

This answer only works under the additional assumption that $V$ is compact and orientable. But I thought it was useful to post nonetheless. I first claim:

Proposition: If $V$ is a compact orientable rational homology $n$-ball, then $\partial V$ is a rational homology $(n-1)$-sphere.

Proof: To see this, consider the LES in rational cohomology for the pair $(V,\partial V)$. A portion looks like $$...\rightarrow H_k(\partial V;\mathbb{Q})\rightarrow H_k(V;\mathbb{Q}) \rightarrow H_k(V,\partial V; \mathbb{Q})\rightarrow ...$$

Since $H_k(V;\mathbb{Q}) = 0$ for $k > 0$, this gives isomorphisms $H_{k+1}(V,\partial V;\mathbb{Q})\cong H_k(\partial V;\mathbb{Q})$ for any $k > 0$. On the other hand, by Poincare-Lefshetz duality (which requires compactness and orientability), $H_{k+1}(V,\partial V;\mathbb{Q})\cong H^{\dim V-(k+1)}(V;\mathbb{Q})$ and the latter group is isomorphic to $H_{\dim V -(k+1)}(V;\mathbb{Q})$ by universal coefficients. Since $V$ is a rational homology sphere, $H_{\dim V - (k+1)}(V;\mathbb{Q}) = 0$ unless $k+1 = \dim V$, when it is isomorphic to $\mathbb{Q}$.

Since $\dim \partial V = \dim V - 1$, we conclude $$H_k(\partial V;\mathbb{Q}) \cong H_{k+1}(V,\partial V;\mathbb{Q})\cong H_{\dim V -(k+1)}(V;\mathbb{Q})\cong \begin{cases} \mathbb{Q} & k=0, \dim \partial V\\ 0 & \text{otherwise}\end{cases}.$$ So, $\partial V$ is a rational sphere. $\square$

Armed with this proposition, for $n\geq 2$, a rational homology sphere $\partial V$ has $H_1(V;\mathbb{Q}) = 0$, so that, in particular, $H_1(V;\mathbb{Z})$ is entirely torsion. Since $\mathbb{Z}$ has no torsion, and my $H_1(V;\mathbb{Z})\rightarrow \mathbb{Z}$ must be trivial. Thus, $Hom(H_1(\partial V;\mathbb{Z}), \mathbb{Z})=0$ as claimed in the paper.