Hi, I'm trying to show that if ## M^n ## is orientable and connected, with boundary (say with just one boundary component), then its top homology is zero. Sorry, I have not done much differential topology/geometry in a while. I'm trying to avoid using Mayer-Vietoris, by using this argument; but I don't think this is too rigorous; please critique: let B be the boundary. Now, B is an (n-1)-submanifold of M, and so it must bound some n-dimensional manifold. Since M is orientable and n-dimensional, then B must bound M itself. But then M, as the top cycle is a cycle that bounds, so the top cycle is a trivial cycle, and so the homology is zero.