A usually hard problem in Algebraic Geometry is computing the fundamental group of the complement $U=\mathbb{P}^n - V$, where $V$ is a reduced hypersurface.

At least in principle, the computation of $\pi_1(U)$ can be carried out by using Seifert-Van Kampen theorem; the tricky part is that the relations in the presentation depend not only on the singularities of $V$, but also on their mutual position.

A classical example due to Zariski is when $V \subset \mathbb{P}^2$ is a curve of degree $6$ having six ordinary cusps and no other singularities. Then $\pi_1(U)$ is the free product $\mathbb{Z}/2 \mathbb{Z} * \mathbb{Z}/3 \mathbb{Z}$ if the six cusps lie on the same conic, and the direct product $\mathbb{Z}/ 2 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z}$ otherwise.

However, by using Lefschetz duality we can compute in a straightforward way $H_1(U, \, \mathbb{Z})$ for all hypersurfaces, so that the abelianization of $\pi_1(U)$ is always known even if a presentation for $\pi_1(U)$ is not available. In fact, there is the following result that can be found in A. Dimca's book *Singularities and topology of hypersurfaces*, Chapter 4.

**Proposition.** Assume that the hypersurface $V \subset \mathbb{P}^n$ has $k$ irreducible components $V_1, \ldots, V_k$ of degrees $d_1, \ldots, d_k$. Let $d$ be the greatest common divisor of the integers $d_1, \ldots, d_k$. Then $$H_1(U, \, \mathbb{Z}) = \mathbb{Z}^{k-1} \oplus \mathbb{Z}/ d \mathbb{Z}.$$

In particular, $H_1(U)$ does not depend on the singularities of $V$, but only on the number and degrees of its irreducible components. As a consequence, we recover the fact that in both cases of Zariski's example quoted above (where $k=1, \, d_1=6$) the abelianization of $\pi_1(U)$ is the cyclic group of order $6$.