Irreducibility of a certain incidence correspondence Let $B_1, B_2, B_3, B_4$ be four irreducible curves in $\mathbb{P}^3$. We can consider the incidence correspondence of lines that meet 4 automorphisms of these 4 curves:
$$
\Phi = \{(\varphi_1, \varphi_2, \varphi_3, \varphi_4, L)\in (\operatorname{PGL_4})^{4}\times \mathbb{G}(1,3)\ | \ L\cap \varphi_{i}(B_i)\neq\emptyset \text{ for each } i\}
$$
How can we show that $\Phi$ is irreducible? I can consider the second projection map $\pi_2:\Phi\to \mathbb{G}(1,3)$. Since $\mathbb{G}(1,3)$ is irreducible, it suffices to show that all the fibers of $\pi_2$ are irreducible. This is not clear to me. I am not sure how to understand the fibers $\pi_{2}^{-1}(L)$ for each $L\in \mathbb{G}(1,3)$.  
This is Exercise 3.31 (page 127) from "3264 and all that..." by Eisenbud and Harris. You can find a free and legal copy of the book from the Harvard website. 
 A: The fiber $\pi_2^{-1}(L)$ is the set of quadruples of automorphisms $(\varphi_1,\varphi_2,\varphi_3,\varphi_4)$ such that $L \cap \varphi_i(B_i)$ is non-empty.
You want to show that this fiber is irreducible.
Note that it is enough to show that $M=\{ \varphi \in PGL_4 \mid L \cap \phi(B) \neq \emptyset \}$ is irreducible, where $B$ is an irreducible curve, since the fiber is the product of four of these.
To prove this, look at another incidence correspondence. Let $\Gamma$ be the set $\{ (\varphi,p) \mid \varphi(p) \in L\} \subset PGL_4 \times B$. Let $\pi_2$ be the projection onto $B$, and $\pi_1$ be the projection onto $PGL_4$. Then $M=\pi_1(\Gamma)$. If we can show that $\Gamma$ is irreducible, we're done.
Let $p \in B$. Then $\pi_2^{-1}(p)$ is the set of automorphisms
$$
\{ \varphi \mid \varphi(p) \in L \} \subset PGL_4
$$
We can for simplicity assume that $L=\{ x_0=x_1=0 \}$. Think of $\varphi$ as a four by four matrix. Then the above equation says that the two first coordinates of $\varphi(p) \in \mathbb P^3$ (thought of as $k^4$), are zero. This is a linear equation in the coordinates on $PGL_4$. Since $GL_4$ is just an open subset of $\mathbb A^{n^2}$, and this defines a linear subspace, it follows that this defines an irreducible set for all $p \in B$. Hence, since $B$ is irreducible, and the map $\pi_2$ have irreducible fibers of the same dimension, it follows that $\Gamma$ is irreducible.
Hence $\pi_1(\Gamma)$ is irreducible, and so is your $M$.
