Let $C \subset \mathbb{C}^n$ be a singular complex curve. Is there a way to compute its (singular) homology? (or at least its betti numbers / Euler characteristic).
If it were non-singular, then $C$ can be viewed as a topological surface, and thus it's homology can be computed using its genus, but singular curves are not topological manifolds.
Can it be computed via its resolution? I know that the genus is the same for $C$ and for the resolution of singularities of $C$, but is the homology preserved as well?
If $C'$ is the normalization of the curve $C$ there are finite sets of points $Z' \subset C'$ and $Z \subset C$ such that $$ C' \setminus Z' \cong C \setminus Z. $$ Now, you can write the excision exact sequences to express the cohomology of $C$ in terms of cohomology of $C'$ (which is a topological manifold) and the finite sets $Z'$ and $Z$.