The complement of an algebraic curve is path-connected

Question

I'm reading a paper on the Zariski-Van Kampen theorem that gives us a presentation of $\pi_1(\mathbb{C}P^2 \setminus K)$ where $K$ is an algebraic curve. The paper uses that $\mathbb{C}P^2 \setminus K$ is path-connected to justify that any point in the set can be chosen as a base point to calculate the fundamental group. I have been thinking about it and I don't know how one could prove that.

Could you give me the idea?

Thank you a lot in advance.

Answer 1

This follows from the well-known duality theorems for manifolds. Let me quote from Hatcher's "Algebraic Topology":

Theorem 3.44. If $K$ is a compact, locally contractible subspace of a closed orientable $n$-manifold $M$, then $H_i(M,M \setminus K; \mathbb Z) ≈ H^{n-i}(K; \mathbb Z)$ for all $i$.

$\mathbb{C}P^2$ is orientable and $K$ is a two-dimensional surface. Thus $H^j(K;\mathbb Z) = 0$ for $j > 2$. This implies $H_0(\mathbb{C}P^2,\mathbb{C}P^2 \setminus K; \mathbb Z) = H_1(\mathbb{C}P^2,\mathbb{C}P^2 \setminus K; \mathbb Z) = 0$. The long exact homology sequence of the pair $(\mathbb{C}P^2,\mathbb{C}P^2 \setminus K)$ shows that $H_0(\mathbb{C}P^2 \setminus K) \to H_0(\mathbb{C}P^2)$ is an isomorphism. Since $H_0(X)$ is a free abelian group whose generators are the path components of $X$, we see that $\mathbb{C}P^2$ and $\mathbb{C}P^2 \setminus K$ have the same number of path components. Thus $\mathbb{C}P^2 \setminus K$ is path connected.