Let $\phi : X \to Y$ be a bijective open morphism of irreducible affine varieties, and suppose that $Y' \subseteq Y$ is an irreducible closed subvariety. Is $X' = \phi^{-1}(Y') \subseteq X$ necessarily irreducible?
I think yes, and here is my reasoning.
Let $\psi : Y \to X$ be the (set) inverse of $\phi$. Then we don't know that $\psi$ is a morphism of varieties, but it is continuous with respect to the Zariski topology on $X$ and $Y$.
Indeed, if $U \subseteq X$ is open, then $\psi^{-1}(U) = \phi(U)$ is open in $Y$ since $\phi$ is an open mapping.
But then $X' = \psi(Y')$ is the continuous image of an irreducible set, and so is irreducible.
Have I got this right?
For context, this issue came up when I was reading the proof of Lemma 5.1.5 in Springer's Linear Algebraic Groups. They have precisely this scenario, but in the proof they consider the cases that $X'$ is reducible or not separately, which makes me confused seeing as it seems to me that $X'$ is always irreducible?