As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper. For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$). This condition is basically that the non-singular $X$ has *rational singularities*. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic [http://front.math.ucdavis.edu/0911.3599][1]. For the second question *"fibres are not unions of projective spaces"*, you've already seen one non-normal example. I believe that fibers of such maps should always have unirational irreducible components (unirational means birational to an image of $\mathbb{A}^n$ under some map) but nothing more. I'm sure one can cook up normal examples too, but I don't know of any with the possible exception of the following counter-example to your third question (which I haven't checked). The last question, *"is it always true that $H^n(Y,O_Y)=0$?"*, here's an example which this time is normal. It is Section III of Cutkosky's [*A new characterization of rational surface singularities*][2] Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in. **Note:** In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this [answer on math overflow][3]. [1]: http://front.math.ucdavis.edu/0911.3599 [2]: http://link.springer.com/article/10.1007/BF01233425 [3]: http://mathoverflow.net/a/43415/3521