Polyhedral homology manifold, but not topological manifold. Let $K$ be a simplicial complex given by a set of vertices $V(K)$ and a distinguished set $S(K)$ (the simplexes) of finite non-empty subsets of $V(K)$. Given a simplex $\sigma \in S(K)$, its link is the subcomplex $Lk(\sigma)$ of $K$ formed by simplexes $\tau \in S(K)$ such that $\tau \cap \sigma=\emptyset$ and $\tau \cup \sigma \in S(K)$. A polyhedral homology $n$-manifold is (according to Hausmann's book) a simplicial complex $M$ such that for each simplex $\sigma \in S(M)$ of dimension $k$, its link $Lk(\sigma)$ is a simplicial complex of dimension $n-k-1$ which has the homology of the sphere $S^{n-k-1}$. Hausmman comments that it can be proven that simplicial complexes whose geometric realization are topological $n$-manifolds are polyhedral homology $n$-manifolds. But I couldn't find a polyhedral homology manifold that is not a topological manifold. Can someone show me one of these? Can it be done in dimension 2?

You can't do this in dimension $2$, but you can do it in dimension $4$. First, start with a triangulated homology $3$-sphere $S$, such as the Poincare homology sphere. Let $M=\Sigma S$. This is a homology manifold but is not a manifold. It is more or less constructed to have the property of a homology manifold. (The link of each cone point is $S$, whereas the links of the other points are suspensions of actual spheres, so are spheres.) Intriguingly, if you suspend twice you get an actual sphere. This is the the Double Suspension Theorem of Cannon and Edwards.