Suppose $X\to Y $ is a morphism , under what conditions we have direct image sheaf $f_*(O_X)=O_Y$?
For example, suppose $\tilde{S}\to S$ is a blow up, do we have $f_*(O_{\tilde{S}})=O_S$?
Hartshorne III 11.3 says that $f_*(O_X)=O_Y$ implies the fibres are connected, is the convese true? That is: if the fibres of a morphism of are all connected, do we have $f_*(O_X)=O_Y$?
Suppose $Y$ is noetherian for simplicity.
For a blowing-up morphism $f: X\to Y$, one has $f_*O_X=O_Y$ if $Y$ is integral and normal. Indeed, $f$ is birational and proper, hence $O_Y\to f_*O_X$ is finite by the direct image theorem, and $f_*O_X$ is contained in the function field of $Y$. So the normality of $Y$ implies that $O_Y=f_*O_X$.
If $Y$ is integral, but not normal, the normalization map $f: X\to Y$ is a blow-up morphism (if $Y$ is a quasi-projective variety), but $f_*O_X$ contains strictly $O_Y$.
A sufficient condition for $f_*O_X=O_Y$ is $f$ proper with geometrically integral fibers. Without properness, the example of Zhen Lin in the comments shows that $f$ can have geometrically integral fiber but still $f_*O_X\ne O_Y$.