maps with connected fibers Let $\pi: X \to S$ be a morphism of schemes. I will say $\pi$ is "pseudoconnected" if $\mathcal{O}_S \to \pi_* \mathcal{O}_X$ is an isomorphism (this is not standard language).


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*If $\pi$ is proper with connected fibers, can we deduce that $\pi$ is pseudoconnected? I think this follows from Zariski's main theorem (the version that says a proper morphism is a pseudoconnected morphism followed by a finite morphism) but I am squeamish because Hartshorne doesn't explicitly say this (even for projective morphisms).

*What if instead $\pi$ is proper with geometrically connected fibers? 
 A: *

*Non. Let $X\to S$ correspond to a finite extension of fields $L/K$ of degree $>1$. Then the unique fiber is connected, but $O_S=K\to \pi_*O_X=L$ is not an isomorphism. 

*Again non. Take $L/K$ purely inseparable in the above example. 
(A sufficient condition when $S$ is reduced is $\pi$ proper with geometrically reduced and geometrically connected fibers.) 
Edit A more geometric example. Let $S$ be a curve over $\mathbb C$ with a cusp and normalization $\rho : S'\to S$. Let $X=\mathbb P^1 \times S'$ with the natural (projective) morphism $\pi$ to $S$. Then $\pi_*(O_X)=\rho_*(O_{S'})\ne O_S$, while the fibers are geometrically connected.
The above sufficient condition is too strong. It implies that for any base change $T\to S$, the direct image of $O_{X\times_S T}$ is equal to $O_T$ (in other words, the isomorphism $O_S\to \pi_*O_X$ holds universally). A weaker condition is the following: suppose that $S$ is Noetherian, integral and normal, that the generic fiber of $X\to S$ is geometrically integral, and $X$ is integral. Then $O_S\to \pi_*O_X$ is an isomorphism. This is because $\mathrm{Spec}(\pi_*O_X)\to S$ is finite and birational. 
