# homology of complex singular curve

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 $$Z$$.
• I guess $Z$ would be the singular locus (finite as $C$ is a curve), but what are $Z'$? Above each point in $Z$ there is an exceptional divisor, which is not finite. Jul 6, 2023 at 11:31
• @SergetheToaster If you construct the normalization explicitly as a sequence of blowups, it will be infinite as a subscheme of the blown up projective space. But the intersection of the exceptional divisor with the normalization $C'$ will of course be finite. Jul 6, 2023 at 12:13