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.



When does a finite morphism induce isomorphism on cohomology?
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.




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EDIT
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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.




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A simple example with curves
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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.


