Let $\pi:X\to S$ be a proper morphism of Noetherian schemes. Is it possible that $\pi\_\*\mathcal{O}\_X\neq \mathcal{O}\_S$ but the natural map $\mathcal{O}\_s^\wedge\to (\pi\_\*\mathcal{O}\_X)\_s^\wedge$ is an isomorphism for all points $s\in S$? The completion is with respect to $\mathfrak{m}\_s$ on both sides.



Completed stalks of the pushforward of the structure sheaf
This has little to do with morphisms, and follows immediately from the following commutative algebra lemma:


**Lemma.** *Let $R$ be a Noetherian ring, and $f \colon M \to N$ a morphism of finite $R$-modules. Then $f$ is an isomorphism if and only if $f^\wedge\_{\mathfrak m} \colon M^\wedge\_{\mathfrak m} \to N^\wedge\_{\mathfrak m}$ is an isomorphism for all maximal ideals $\mathfrak m \subseteq R$.*


*Proof.* There are many ways to prove this. For example, you can use that $R\_\mathfrak m \to R^\wedge\_\mathfrak m$ is faithfully flat (Tag [00MC](https://stacks.math.columbia.edu/tag/00MC)), and for finite $R\_{\mathfrak m}$-modules $P$ we have $P^\wedge \_{\mathfrak m} = P \otimes\_{R\_{\mathfrak m}} R^\wedge\_{\mathfrak m}$ (Tag [00MA](https://stacks.math.columbia.edu/tag/00MA)). Thus a finite $R\_\mathfrak m$-module $P$ is zero if and only if $P ^\wedge\_{\mathfrak m} = 0$ (this also follows from Nakayama's lemma), and $f^\wedge\_{\mathfrak m}$ is an isomorphism if and only if $f\_{\mathfrak m} \colon M\_{\mathfrak m} \to N\_{\mathfrak m}$ is. Now the result follows from Tag [00HN](https://stacks.math.columbia.edu/tag/00HN). $\square$


