Why isn't $H_n(M) \cong 0$ for any $n$-manifold $M$ with non-trivial boundary?

Question

I learned here that if we have a connected "open manifold" $M$ of dimension $n$, then $H_n(M) \cong 0$. I'm not quite sure why this can't be used to prove that every connected $n$-manifold with non-trivial boundary has $H_n(M) \cong 0$. In particular, doesn't this mean that $M\setminus \partial M$ has trivial $n$th homology group? If so, does this imply that $M$ is not homotopy equivalent to $M \setminus \partial M$, and if not, why isn't $H_n(M)$ trivial for any $n$-manifold $M$ with non-trivial boundary?

Answer 1

Indeed, for e.g singular homology, if $M$ is connected and with boundary we have $H_n(M) = 0$.

However, let me quickly mention Borel-Moore homology. The theory was made to extend Poincaré duality to more general spaces. There is always a top class in Borel-Moore homology, for example $H^{BM}_i(\Bbb R^n) = \Bbb Z$ if $i=n$ and $0$ else (notice that it is not homotopy invariant).