2

I am doing a problem which ask to show the following:

Assume we have a 3-dimensional closed orientable manifold M with fundamental group $Z_p$, then M is not homeomorphic to any subspace of $Y$ of $S^4$ such that the complement space of $Y$ in $S^4$ is disjoint union of path-connected space $A$ and $B$, and they have neibourhood $N(A)$ and $N(B)$ such that $N(A)$ and $N(B)$ are homotopic equivalent to $A$ and $B$, intersection of $N(A)$ and $N(B)$ is homotopic equivalent to $Y$.

The hint is to show $H_1(A)\oplus H_1(B)=H_1(M)$ and $H_1(A)=H_1(B)$. For $H_1(A)\oplus H_1(B)=H_1(M)$ I believe we can use Alexander duality. What really confuses me is to show $H_1(A)=H_1(B)$, I have tried all the tools I know but have no idea. Is there any intuition for this to be true? I appreciate any proof for this!

user330928
  • 826
  • This is somewhat nontrivial and the only proof I know involves the torsion linking form (it somehow must, because this is certainly false for groups that aren't finite). Is that machinery you know? –  Aug 24 '16 at 06:27
  • @MikeMiller Unfortunately nope. Actually I see this problem from a qualifying exam...... – user330928 Aug 24 '16 at 14:38

0 Answers0