If $f:X \to Y$ is a flat and proper surjective morphism between smooth schemes over an algebraically closed field, and $f$ has connected fibers, does it imply that $$f`_*\mathcal O_X = \mathcal O_Y?$$

This follows from Zariski's main theorem if the characteristic is zero and it is false in positive characteristics: consider the the morphism $\mathbb{A}^1 \to \mathbb{A}^1$ given by $x \mapsto x^p$ where $p$ is the characteristic. The statement would also be true in char p if you assume that the general fibre is reduced. (Note that it suffices to assume that X is integral and Y is normal. $f$ should of course also be surjective.) 


If $X$ is smooth over $K$ and if $K$ is of characteristic $p>0$, then the relative Frobenius $F:X\rightarrow X\times_{F_X} K=:X'$ is faithfully flat $K$morphism and finite, hence proper. Moreover it's a homeomorphism, so the fibers are connected. But it's not hard to see that $F_*\mathcal{O}_{X}$ is locally free of rank $p^n$, if $X$ is of dimension $n$. 


Since $f : X \to Y$ is a projective morphism, we can use Stein factorization and write $f = h \circ g$ with
Since $f$ has connected fiber and surjective, $h$ then must be surjective and has degree 1. It follows that $h$ is an isomorphism. Hence, $f_*\mathcal{O}_X = \mathcal{O}_Y$. I think we only need to assume that $X$ and $Y$ are noetherian, normal and over $\mathbb{C}$. 


Over $\mathbb{C}$, I would reason as follows. For any $V$ open in $Y$, $\mathcal{O}_Y(V) \cong \mathcal{O}_X(U)$, where $U:=f^{1}(V)$, via the pullback $f^*:\mathcal{O}_Y (V) \rightarrow \mathcal{O}_X (U)$ $f^*:h \mapsto h \circ f$, thinking of $h$ as a morphism $h:Y \rightarrow \mathbb{A}^1$. The above pullback is in general injective, but in this case it's even surjective because every regular function $g$ on $X$ has constant value on the fibers (by base change, in this case the (reduction of the) fibers are connected proper varieties), hence is the pullback of a function $h$ on $Y$. 

