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I want to calculate the fundamental group of the complement some collection of plane curves (specifically, two nodal cubics in a general position).

I've read about Severi problem (solved by Harris), which states that complement of every reduced irreducible nodal plane curve has an abelian fundamental group. I've understood the idea of the argument (deforming the picture to the arrangement of lines, which only increases the fundamental group).

I think the same (the fact that the complement has an abelian fundamental group) should hold without irreducibility assumption (i.e., for an arrangement of nodal curves which also intersect transversely). Is it true? Is it known?

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    $\begingroup$ The result is actually due to Fulton (On the fundamental group of the complement of a nodal curve, Ann. of Math. 111, no. 2(1980), 407-409), and it is valid with no irreducibility assumption. $\endgroup$ – abx May 13 at 18:27
  • $\begingroup$ @abx Thank you! I would accept this is as an answer. $\endgroup$ – Lev Soukhanov May 13 at 20:34
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As suggested, I write my comment as an answer: the result is actually due to Fulton (On the fundamental group of the complement of a nodal curve, Ann. of Math. 111, no. 2 (1980), 407-409), and it is valid with no irreducibility assumption. Fulton works with the algebraic fundamental group; the case of the topological fundamental group was done by Deligne (Séminaire Bourbaki 1979/80, Exp.543, pp. 1-10.

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