If $M$ is a connected, non-compact $n$-manifold, then $H_i(M;R)=0$ for $i\geq n$. For a proof, see Proposition 3.29 in Hatcher's Algebraic Topology book.

So, if you are going to have $H_n(M;R)=R$, $M$ had better be compact.

If $M$ is a connected, non-compact $n$-manifold, then $H_i(M;R)=0$ for $i\geq n$. For a proof, see Proposition 3.29 in Hatcher's Algebraic Topology book.

So, if you are going to have $H_n(M;R)=R$, $M$ had better be compact.

EDIT (to answer about homology $n$-manifolds):

A homology $n$-manifold is a finite dimensional, locally contractible space $X$ whose local homology groups $H_*(X, X-\{x\})$ are the local homology groups for $\mathbb{R}^n$ for every $x\in X$. In particular, $\mathbb{R}^n$ is a non-compact homology $n$-manifold.

If $M$ is a connected, non-compact $n$-manifold, then $H_i(M;R)=0$ for $i\geq n$. For a proof, see Proposition 3.29 in Hatcher's Algebraic Topology book.

So, you you are going to have $H_n(M;R)=R$, $M$ had better be compact.

If $M$ is a connected, non-compact $n$-manifold, then $H_i(M;R)=0$ for $i\geq n$. For a proof, see Proposition 3.29 in Hatcher's Algebraic Topology book.

So, if you are going to have $H_n(M;R)=R$, $M$ had better be compact.