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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.

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  • $\begingroup$ You may want to assume $Y$ is connected or else a simple counter example exists. $\endgroup$
    – Rocket Man
    Jun 3 '18 at 21:03
  • $\begingroup$ @RocketMan Yes thanks for that! $\endgroup$
    – user101010
    Jun 3 '18 at 21:05
  • $\begingroup$ Is the 4-manifold required to be smooth, or can it just be a topological 4-manifold? This might make a big difference. $\endgroup$
    – Carl
    Jun 7 '18 at 3:08
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$\require{AMScd}$

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.

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  • $\begingroup$ This is a great answer. But it's a bit confusing to jump to $H^1$, and then back to $H_1$. It's enough for what you want to do that $H^1(X)\rightarrow H^1(T^3)$ is zero. $\endgroup$
    – Steve D
    Jun 14 '18 at 23:02

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