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.
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)
I just found out this is a simple application of Lefschetz Duality. (See Hatcher, Chap. 3 Sec.3 p.254 as a special case of theorem 3.43) By this duality, $H_n(M)=H^0(M,\partial M)=0$.
