Does the 3-manifold $S^1\times S^2$ bound a smooth integral homology ball?

Question

Does the 3-manifold $S^1\times S^2$ bound a smooth (integral) homology ball?

The only 4-manifolds I know whose boundary is $S^1\times S^2$ are $S^1\times D^3$ and $D^2\times S^2$, and both are not homology balls.

Answer 1

Let $M$ be a $n$-dimensional homology ball. As $M$ is orientable, Lefschetz duality applies, so we have $H^k(M, \partial M) \cong H_{n-k}(M)$. Since $M$ is a homology ball, we have $H^n(M, \partial M) \cong \mathbb{Z}$ and $H^k(M,\partial M) = 0$ for $k \neq n$.

On the other hand, there is a long exact sequence of the pair $(M, \partial M)$

$$\dots \to H^{k-1}(\partial M) \to H^k(M, \partial M) \to H^k(M) \to H^k(\partial M) \to H^{k+1}(M, \partial M) \to \dots$$

Again, as $M$ is a homology ball, we see that $H^k(M, \partial M) \cong H^{k-1}(\partial M)$ for $k > 1$. As the map $H^0(M) \to H^0(\partial M)$ induced by restriction is an isomorphism, we see that $H^1(M, \partial M) = 0$ and $H^0(M, \partial M) = 0$.

Combining these two observations, we see that

$$H^k(\partial M) \cong \begin{cases} \mathbb{Z} & k = 0, n - 1\\ 0 & \text{otherwise}. \end{cases}$$

Therefore, we have arrived at the following conclusion:

If $M$ is a homology $n$-ball, then $\partial M$ is a homology $(n-1)$-sphere.

It follows that $S^1\times S^2$ cannot bound a homology ball as it is not a homology sphere.