Is there any connected n-manifold such that $H_n(X,Z)=Z\times Z$?

Question

I think the question is equal to whether a n-manifold has a n-submanifold which is compact in n-manifold. I feel there is not such manifold, but I don't know how to prove it. In fact, I just need some reference about the proof. Thank you.

Answer 1

If you assume the manifold to be connected, the answer is not, there aren't such manifolds. For any closed $n$-manifold $M$, we have $H_n(M; \Bbb Z) = \Bbb Z$ if $M$ is orientable, or $H_n(M; \Bbb Z) = 0$ if it is not orientable. If $M$ is not closed then $H_n(M; \Bbb Z) = 0$. This is definitely a special case of Poincar\'e duality, although it follows from the existence of a handlebody decomposition with only one $n$-handle for closed $M$, or without $n$-handles otherwise. You can look at almost any book about homology theory (e.g. Vick), or about differential topology. Of course, in the disconnected case you can easily construct manifolds which satisfy your condition (take the disjoint union of any two closed, orientable, connected $n$-manifolds), see other comments.