If $f: X \rightarrow Y$ is a proper morphism of locally noetherian schemes with $f_* \mathcal{O}_X = \mathcal{O}_Y$ then the thm. of formal functions tells us that $f$ has connected fibers, since there is an isomorphism $$ \widehat{ (f_* \mathcal{O}_{X,y})} \longrightarrow \lim_{\leftarrow_n} H^0(X_n, \mathcal{O}_{X_n})$$ for any $y \in Y$, where LHS is the stalk of the pushforward completed as an $O_y$ module and $X_n$ denotes the $n$th thickened fiber as usual.

$\textbf{Question:}$ In general, for coherent sheaves on $X$ the theorem of formal functions only gives an isomorphism of $\widehat{\mathcal{O}_y}$ modules. In the context of Stein factorization above, since both LHS and RHS are in fact rings, is the isomorphism of the thm. of formal functions "sufficiently functorial" so that it is in fact an isomorphism of rings? Then we know that the spectrum of RHS is connected because LHS is a local ring. This seems to be the argument in Illusie's article "Grothendieck existence thm...", but I can't help but think that the thm of formal functions only gives an isomorphism of $\widehat{\mathcal{O}_y}$ modules, so to be completely precise we need a slightly messier argument. Can you make this a general statement, eg, the thm. of formal functions always gives a map of rings we start with a coherent sheaf of rings on $X$.

definitionof the map in the theorem on formal functions (i.e., don't treat the existence of that isomorphism as a black box: the map is defined in a specific way prior to the statement of the result), and then the affirmative answer should become clear to you. Hint: enough to check the ring property after composing with the map from the inverse limit to each of the constituents of the inverse system. $\endgroup$ – user27056 Nov 3 '12 at 17:45`$g^*f_* \rightarrow f'_*g'^*$`

on cohomology explicitly by Cech cohomology. Is this what you had in mind? $\endgroup$ – LMN Nov 3 '12 at 19:16