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 $\mathbb{Z}^{\wedge}_pC(x,y)$?
Best Answer
It suffices to show that you can lift isomorphisms along $AC\to BC$ whenever $A\to B$ is a square zero extension. (EDIT : here I'm using finite generation of $C(y,x), C(x,y)$ to obtain that $\mathbb Z_p\otimes C(x,y)$ is $p$-adically complete)
So let $\sum_i b_i\otimes f_i \in BC(x,y)$ be an isomorphism, with inverse $\sum_j d_j\otimes g_j$.
Choose lifts $\tilde b_i, \tilde d_j$ in $A$ of $b_i, d_j\in B $. Then $(\sum_i \tilde b_i\otimes f_i)\circ (\sum_j \tilde d_j \otimes g_j) = id_y + \epsilon$, where $\epsilon \in \ker\otimes C(y,y)$, where $\ker$ is the kernel of $A\to B$. Indeed, because $C(y,y)$ is flat, $\ker\otimes C(y,y)$ is in particular the kernel of $A\otimes C(y,y)\to B\otimes C(y,y)$ (EDIT: as R. van Dobben de Bruyn pointed out, I'm not using flatness here, just right exactness of the tensor product).
In particular, because $\ker^2 = 0$, you find that $\epsilon\circ \epsilon = 0$, and so $id_y + \epsilon$ is invertible in $A\otimes C(y,y)$. In particular, $\sum_i \tilde b_i \otimes f_i$ has a right inverse, and $\sum_j \tilde d_j \otimes g_j$ has a left inverse.
Now (with the same lifts !) reasoning symmetrically (in $C(x,x)$ ) shows that they each have an inverse on the other side, so they are both isomorphisms. Either one of them is a lift, as desired.
In particular, if you were doing this over another base ring than $\mathbb Z$, really the only thing you would need is for an analogue of $\mathbb Z_p\otimes C(x,y)$ (resp. $(y,x)$) being $p$-adically complete.