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
$$f`_*\mathcal O_X = \mathcal O_Y?$$
[Math] Is the direct image of the structure sheaf on X isomorphic to the structure sheaf on Y when X->Y is flat and proper with connected fibers between smooth schemes over an algebraically closed field
ag.algebraic-geometry
Related Solutions
I don't think that map is always surjective. For example, suppose that $Y = X$ is a supersingular elliptic curve. Then the Frobenius map $H^1(X, O_X) \to H^1(Y, O_Y)$ is the zero map (and both are 1-dimensional vector spaces). See for example Hartshorne's chapter on elliptic curves.
Alternately, suppose that $X$ is a scheme over $\mathbb{R}$, and $Y = X \times_{\mathbb{R}} \mathbb{C}$ is the base change. I don't think one should expect that $H^n(X, O_X) \to H^n(Y, O_Y)$ is basically ever surjective unless they are both zero...
With regards to your question though, here's one answer:
Suppose that $X$ is a normal integral scheme of characteristic zero and $Y$ is also integral. Then the natural map $O_X \to f_* O_Y$ splits as a map of $O_X$-modules, say with splitting map $\phi : f_* O_Y \to O_X$ (use the trace map on the fields $K(Y) \to K(X)$ and restrict to the structure sheaves). Now apply the functor $H^n(X, \bullet)$ to the composition (which is an isomorphism): $$ O_X \to f_* O_Y \xrightarrow{\phi} O_X. $$ Clearly one gets that $$ H^n(X, O_X) \to H^n(X, f_* O_Y) = H^n(Y, O_Y) \to H^n(X, O_X) $$ is also an isomorphism and thus $$ H^n(X, O_X) \hookrightarrow H^n(Y, O_Y) $$ injects as desired.
EDIT
Since the author of the question is particularly interested in the case when $f : Y \to X$ is the normalization of an $n$-dimensional $X$, let me try to say a couple things about that case.
Since we have a short exact sequence $0 \to O_X \to f_* O_Y \to C \to 0$, and since $f$ is birational, the support of $C$ has dimension $< \dim X$. Therefore, $H^n(X, O_X) \to H^n(Y, O_Y)$ is surjective as the original question states.
To show injectivity, it is sufficient to show that $H^{n-1}(X, C) = 0$. This will happen certainly if the non-normal locus of $X$ has codimension $\geq 2$. Otherwise, it generally won't happen.
A simple example with curves
Suppose that $X$ is a curve with exactly a node and $Y$ is its normalization (although any singular curve will work). Then $C$ is the skyscraper sheaf supported at a point. In particular $\dim H^0(X, C) = 1$. On the other hand, $\dim H^0(Y, O_Y) = 1 = \dim H^0(X, O_X)$, and so the exact sequence $$ 0 \to H^0(X, O_X) \to H^0(Y, O_Y) \to H^0(X, C) \to H^1(X, O_X) \to H^1(Y, O_Y) \to 0 $$ immediately implies that $H^0(X, C) = \ker H^1(X, O_X) \to H^1(Y, O_Y)$. In particular, the latter map is not injective.
There is a very general criterion for a map on $\pi_1$ to be surjective. Recall that for $X$ connected, the category of finite étale covers of $X$ is equivalent to the category $\pi_1(X)\text{ -}\operatorname{Set}_f$ of finite sets with a continuous $\pi_1(X)$-action. Under this correspondence, the $Y \to X$ finite étale with $Y$ connected correspond to the connected $\pi_1(X)$-sets $S$ (i.e. $\pi_1(X)$ acts transitively on $S$).
Lemma. Assume $X$, $Y$ connected, and $f \colon X \to Y$ a morphism. Then the induced morphism $\pi_1(f) \colon \pi_1(X) \to \pi_1(Y)$ is surjective if and only if for every $Z \to Y$ finite étale with $Z$ connected, the pullback $Z_X \to X$ is connected.
Proof. If $\pi_1(f)$ is surjective, then clearly any connected $\pi_1(Y)$-set is connected as $\pi_1(X)$-set. Conversely, if $\pi_1(f)$ is not surjective, then some $\gamma \in \pi_1(Y)$ is not in the image. Since fundamental groups are profinite, the image of $\pi_1(f)$ is closed, so the image of $\pi_1(f)$ misses some open neighbourhood of $\gamma$. Thus, there exists an open subgroup $U \subseteq \pi_1(Y)$ such that $$\gamma U \cap \operatorname{im} \pi_1(f) = \varnothing.$$ Then the finite $\pi_1(Y)$-set $S = \pi_1(Y)/U$ is not connected as $\pi_1(X)$-set. But it is clearly connected as $\pi_1(Y)$-set. $\square$
To apply this to the specific geometric setting you are interested in, just note that if $f \colon X \to Y$ has connected geometric fibres, then the same holds for the base change to any finite étale covering $Z \to Y$. It is then clear that if $Z$ is connected, so is $Z \times_Y X$.
Remark. There are more equivalent criteria for surjectivity; see for example Tag 0B6N. The one I gave above is amongst the ones listed, but this was not the case at the time of writing; hence my writing out the proof. My proof above was originally part of the proof of Tag 0BTX.
Best Answer
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.)