Algebraic Curves Lecture 16:

Proof.   Introduce the sets

and Note that passing from $\tilde I(2)$ to $I(2)$ simply requires the plane to meet the curve transversely. Now let $\pi_1:\tilde I(2)\to (C\times C)\setminus \Delta$ be the projection map (here we remove the diagonal since we required that $p_1\neq p_2$ for a tuple in $\tilde I(2)$.

The set $\pi_1^{-1}(p_1,p_2)$ is the set of all hyperplanes passing through $p_1$ and $p_2$, and it is isomorphic to $\mathbb P^{r-2}$. Claim: both $\tilde I(2)$ and $I(2)$ are both irreducible and are in particular connected. > **Side Remark:** if $f:X\to Y$ is some dominant map of varieties and you know that $X$ is irreducible, you can immediately conclude that $Y$ is irreducible. However, if $Y$ is irreducible, concluding that $X$ is irreducible is much harder. A sufficient condition is the following: (1) $Y$ is irreducible (2) the fibers of $f$ are all irreducible (3) the fibers have the same dimension and (4) $f$ is proper. The space $C\times C\setminus \Delta$ is irreducible -- the diagonal is a codimension 2 subset in this case. Furthermore, $\pi_1$ satisfies (1)-(4) in the remark above, so $\tilde I(2)$ is irreducible. A similar argument tells me that $I(2)$ is irreducible. Now we return our attention to the map and fiber

This means $\pi_1(U,H_0)$ acts 2-transitively on $\pi^{-1}_2(H_0)$. Let $G\subseteq S_d$ be the transitive subgroup of $S_d$ given as the image of the fundamental group $\pi_1(U,H_0)\to \operatorname{Aut}(\pi_2^{-1}(H_0)) = S_d$. To show that $G = S_d$ it suffices to show that $G$ contains a transposition. Indeed, if it has a transposition $\tau \in G$, then

1. I can get any other transposition $\tau'$ by conjugation with a permutation $\sigma$ such that $(\tau_1,\tau_2) \mapsto (\tau_1', \tau_2')$. Then $\sigma\tau\sigma^{-1} = \tau'.$

2. The transpositions generate $S_d$, so $G = S_d$.