As stated in the title, my question is the following:
Let $M$ be a compact orientable manifold with boundary $\partial M$. Is it true that $H_n(M;\mathbb{R})$ is always zero?
In the trivial case for compact surfaces in $\mathbb{R}^3$, filling up the interior always gives something homotopically equivalent to an object of lesser dimension. However, I don't see a way of using this method for the general case. Manipulation with Stokes theorem doesn't seem to work either.
Best Answer
If your manifold is triangulated then if you want to find an $n$-cycle $c$, it is a linear combination of the $n$-simplices $\sigma_i$, $c=\sum a_i \sigma_i$. As in the case of a closed manifold, $\partial c=0$ implies that $a_i$'s are all equal. But it you have a boundary, you also get that $a_i=0$ for every $\sigma_i$ which is at the boundary. (I suppose that $M$ is connected, otherwise one needs to consider every component separately)