It's from Vakil's FOAG, exercise 25.1.H, with the following text:
If a morphism $\pi: X \rightarrow Y$ satisfies the property that the natural map $\mathscr{O}_Y \rightarrow$ $\pi_* \mathscr{O}_{\mathrm{X}}$ is an isomorphism, we say that $\pi$ is $\mathscr{O}$-connected. Clearly the notion of $\mathscr{O}$ connectedness is local on the target, and preserved by composition.
25.1.H. EASY EXERCISE. Show that proper $\mathscr{O}$-connected morphisms of locally Noetherian schemes are surjective.
I got some hint that it can be proved via the base change theorem of coherent sheaves. But I cannot show that $\mathcal{O}_X$ is flat over $Y$, i.e., the morphism is flat, which is required by the base change theorem.
Is that the condition flat is missed? I saw the following result A simple geometric proof of Zariski's connectedness theorem? too, but I found that the empty set is also connected. Hence it does not help here.
I got it from Stacks#03GY, a lemma used for the Stein factorization. The Argument is copied as follow:
Say $\pi: X \to Y$ is proper, then it's closed, and quasicompact. WLOG suppose $Y = \operatorname{Spec}B$. Since $\pi$ is closed the image of $X$ has the form $V(I)$. For any element $h \in I$, the pullback of it to $\mathcal{O}(X)$, denoted as $\pi^\sharp h$ vanishes at all point of $X$. Now since $\pi$ is quasicompact, $X$ is quasicompact, hence there is some $n > 0$ such that $(\pi^\sharp h)^n = 0$. But from the assumption $\pi$ being $\mathcal{O}$-connected, $\mathcal{O}(X)$ is isomorphic to $B$. Hence $h^n = 0$. Hence $I$ is contained in the nilradical of $B$, and $V(I)= \operatorname{Spec}B$.
From the argument it shows closed, quasicompact and $\mathcal{O}$-connected morphisms of schemes are surjective. Other conditions are unnecessary.