Assume that $X$ is the complement of a plane algebraic curve $C$ in $\mathbb{C}^2$ and Y is the complement of the union of $C$ and a line $L$ (not contained in $C$). Assume that $Y$ is $K(\pi, 1)$. Is it true that $X$ is $K(\pi, 1)$? Why or why not? 



Homotopy type of complement of a plane algebraic curves.
The answer to this question is no. Let me explain why, the explanation uses two facts.


**1)** The complement to the collection of $4$ generic lines in $\mathbb CP^2$ is not $K(\pi,1)$.


Indeed, it is not hard to see that the fundamental group of this complement is $\mathbb Z^3$. At the same time the complement has homotopy type of a $CW$ complex of dimension $2$ and if it were $K(\pi, 1)$, it would zero cohomology in degree $3$, while $H^3(\mathbb Z^3)=\mathbb Z$. 


**2)** There exist line arrangements in $\mathbb CP^2$ that contain $4$ generic lines, but at the same time their complement is $K(\pi,1)$. For example, you can take the arrangement given by $6$ lines $x=0$, $y=0$, $z=0$, $x=y$, $y=z$, $z=x$. The generic four lines in it are $x=0$, $y=0$, $x=z$, $y=z$.


**Proof for No.** Taking the above arrangement throw away from first the line $z=0$, second the line $x=y$. Hence at the first or second time we will pass from an arrangement whose complement is $K(\pi,1)$ to the one whose complement is not $K(\pi,1)$.
**End**. 


You can check that the complement to the above $6$ lines arrangement is $K(\pi,1)$ either directly, or you can use, for example the theorem of Deligne, that tells us that the complement to a complexification of a real simplicial arrangement is $K(\pi,1)$. (in our case real simplicial means that the $6$ lines are real and they cut $\mathbb RP^2$ in triangles.)


Here is the reference: Deligne, Pierre Les immeubles des groupes de tresses généralisés. (French) Invent. Math. 17 (1972)


