The Stacks project

Here is a proof:

Suppose $f(E)$ is the union of $Z_1 \cap f(E)$ and $Z_2 \cap f(E)$, for two distinct closed subsets $Z_1$ and $Z_2$ of $Y$; this is equal to the intersection $(Z_1 \cup Z_2) \cap f(E)$, so $f(E)$ is then contained in the union $Z_1 \cup Z_2$. For the irreducibility of $f(E)$ it suffices to show that it is contained in either $Z_1$ or $Z_2$. The relation $f(E) \subseteq Z_1 \cup Z_2$ shows that $f^{-1}(f(E)) \subseteq f^{-1}(Z_1 \cup Z_2)$; as the right-hand side is clearly equal to $f^{-1}(Z_1) \cup f^{-1}(Z_2)$ and since $E \subseteq f^{-1}(f(E))$, it follows that $E \subseteq f^{-1}(Z_1) \cup f^{-1}(Z_2)$, from which one concludes by the irreducibility of $E$ that $E \subseteq f^{-1}(Z_1)$ or $E \subseteq f^{-1}(Z_2)$. Hence one sees that either $f(E) \subseteq f(f^{-1}(Z_1)) \subseteq Z_1$ or $f(E) \subseteq Z_2$.

Added the proof and fix long line. Thanks!

Things get simpler if you first replace $f$ by $E\to f(E)$. The statement becomes: "if $X$ is irreducible and $f$ surjective then $Y$ is irreducible". Now of course $Y$ is nonempty if $X$ is, and then if $Y=Y_1\cup Y_2$ with $Y_1$, $Y_2$ closed, then from the irreducibility assumption you get that $X=f^{-1}(Y_i)$ for some $i$, hence $Y=Y_i$ since $f$ is surjective.

OK, yes what you say makes sense, but since the proof is correct and since this is so early in the theory, it isn't worth it to me to make the change. If you want to make the edits to the latex file and send it to me, then I'll reconsider.