Let $Y$ be a connected closed orientable 3-manifold. Is there a compact orientable 4-manifold $X$ with $\partial X = Y$ such that the map induced by the inclusion $i_*: H_2(Y; \mathbb{Z}) \to H_2(X; \mathbb{Z})$ is the zero map.
The best I can do so far is to show this in the case where the cup product $\cup: H^1(Y; \mathbb{Z}) \times H^1(Y; \mathbb{Z}) \to H^2(Y; \mathbb{Z})$ vanishes.
$\require{AMScd}$
I will drop the coefficients and use $\mathbb{Z}$ throughout.
For $T^3$, for example, there is no such $X$. Suppose that there were. If $H_2(T^3) \to H_2(X)$ is 0, then $H_1(T^3) \to H_1(X)$ will also be 0, since the cup product $\cup : H^1(T^3) \times H^1(T^3) \to H^2(T^3)$ is surjective. By considering the relative long exact sequence, the map $H_3(X,T^3) \to H_2(T^3)$ will be surjective. By duality, we then have the commutative diagram:
\begin{CD} H_3(X, T^3) @>{}>> H_2(T^3)\\ @VVV @VVV\\ H^1(X) @>{}>> H^1(T^3)\\ @VVV @VVV\\ \text{Hom} (H_1(X),\mathbb{Z}) @>{0}>> \text{Hom}(H_1(T^3), \mathbb{Z}) \end{CD}
and therefore $H_1(T^3) = 0$ which is a contradiction.