Let $Y \to X$ be a finite morphism of schemes of dimension $n$. The induced map on the top cohomology $H^n(X, \mathcal O_X) \to H^n(Y, \mathcal O_Y)$ is always surjective. When is it injective? Can one achieve this by adding reasonable conditions? We may restrict ourselves to projective varieties over an algebraically closed field.
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 1dimensional 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 $H^{n1}(X, C) = 0$. This will happen certainly if the nonnormal 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.

1$\begingroup$ Using the flat base change theorem, we can formalize your "I don't think one should expect that $H^n(X,\mathcal O_X) \to H^n(Y,\mathcal O_Y)$. Because $\mathbb C$ is flat over $\mathbb R$, $Y\to X$ is flat base change, so $H^n(Y,\mathcal O_Y)=H^n(X,\matcal O_X) \otimes_{\mathbb R} \mathbb C$, and the map between them is the obvious one. It is indeed never surjective when $H^n(X,\mathcal O_X)\neq 0$. $\endgroup$ – Will Sawin Sep 12 '12 at 17:08


$\begingroup$ @Karl Scwede: Thanks for the answer. I see that the map in question need not be surjective. I had the normalization map $\bar{X} \to X$ in mind. Can we achieve injectivity if $X$ is not normal? Can the trace map be defined in nonnormal situations? $\endgroup$ – Paul Graaf Sep 14 '12 at 4:29

$\begingroup$ Definitely you can't do the trick I did above. There is never a splitting of $O_X \to O_{\widetilde{X}}$ and the trace is worthless. Let me think about normality briefly and then edit my answer. $\endgroup$ – Karl Schwede Sep 14 '12 at 11:33