TITLE: Lifting isomorphisms between linear categories
QUESTION [8 upvotes]: Let $C$ be a $\mathbb{Z}$-linear category, such that $C(x,y)$ is a free abelian group with finite rank, for every $x,y\in\mathrm
{Ob}(C)$. Given a commutative ring with identity $R$, let $RC$ denote the category with the same objects of $C$, and morphisms $RC(x,y):=R\otimes_{\mathbb{Z}} C(x,y)$.
Does any isomorphism in $\mathbb{F}_pC(x,y)$ lift to an isomorphism in $\mathbb{Z}^{\wedge}_pC(x,y)$?

REPLY [4 votes]: Here's a simplification of Maxime Ramzi's answer. Given an isomorphism $\bar f \in \mathbf F_p\mathscr C(x,y)$, the claim is that any lift $f \in \mathbf Z_p\mathscr C(x,y)$ of $\bar f$ is an isomorphism.
Indeed, if $g \in \mathbf Z_p\mathscr C(y,x)$ is a lift of $\bar g = \big(\bar f\big)^{-1}$, we know that $gf = 1 - p\phi$ for some $\phi \in \mathbf Z_p\mathscr C(x,x)$. The series $\psi = \sum_{i=0}^\infty p^i\phi^i$ converges in $\mathbf Z_p\mathscr C(x,x)$ and gives a two-sided inverse of $gf$, so $f$ has a left inverse $\psi g$ (and $g$ has a right inverse $f\psi$). Swapping the roles of $f$ and $g$ shows that $f$ has a right inverse as well, so $f$ is invertible. $\square$

Remark. A note on the hypotheses: flatness is never used; the only role it plays in Ramzi's answer is uniqueness of $\phi$ such that $gf = 1-p\phi$, but this is inconsequential. On the other hand, it is crucial that the $\mathscr C(x,y)$ are finitely generated, as we need $\mathbf Z_p\mathscr C(x,y)$ to be $p$-adically complete [Tag 00MA(3)].