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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.