TITLE: Killing 2nd homology by including 3-manifold as the boundary of a 4-manifold
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: $\require{AMScd}$
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.