This is a problem concerning a lemma in Oka's paper "On the fundamental group of the complement of a reduced curve in $\mathbb{P}^2$". Let $C$ be a curve in $\mathbb{P}^2$ and $L$ be a general line to $C$. The lemma says that $\pi_1(\mathbb{P}^2-C\cup L)$ is abelian if and only $\pi_1(\mathbb{P}^2-C)$ is abelian. Now consider the arrangement of a conic $C$ and three lines $L_1$, $L_2$ and $L_3$ in $\mathbb{P}^2$. Assume that the conic passes through the three double points $L_1\cap L_2$, $L_1\cap L_3$ and $L_2\cap L_3$. We can show that $\pi_1(\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3)$ is abelian. Let $L_\infty$ be a general line to the arrangement. According to Oka's lemma, the fundamental group $\pi_1(\mathbb{C}^2-C\cup L_1\cup L_2\cup L_3)=\pi_1(\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3\cup L_\infty)$ should be abelian. However, without the projective relation, it is impossible to prove that the group is abelian. I don't know where I made mistakes. The only mistake that I suspect is that $\mathbb{P}^2-C\cup L_1\cup L_2\cup L_3\cup L_\infty$ does not equal $\mathbb{C}^2-C\cup L_1\cup L_2\cup L_3$. But why they do not equal.