Does linearization of categories reflect isomorphism?

Hi Tilman. I believe I proved that (in your language) linearization reflects isomorphism. The following is a sketch. I will send you a more detailed version. The general case may be reduced to the case of prime fields $F_p$ and certain categories $C$ with fixed objects $x$ and $y$ and morphisms $f_1,\dots,f_m\colon x\to y$ and $g_1,\dots,g_n\colon y\to x$ subject to relations which correspond to the fact that $u=f_1+\dots+f_m$ and $u^{-1}=g_1+\dots+g_n$ are mutually inverse in the $F_p$-linearization. Apart from trivial cases, we may reindex these generators such that $f_1g_1 = 1_y$ and $g_nf_m=1_x$, while the other summands in the expansion of $uu^{-1}$ and $u^{-1}u$, respectively, fall into equivalence classes whose size is a multiple of $p$. It is then possible to derive a sequence of pairs $(i_1,j_1),(i_2,j_2),\dots,(i_k,j_k)$ such that $f_{i_r}g_{j_r} = f_{i_{r+1}}g_{j_{r+1}}$ for $r=2,3,\dots,k-1$ and $g_{j_r}f_{i_r}=g_{j_{r+1}}f_{i_{r+1}}$ for $r=1,3,\dots,k-2$. Then $f_{i_1}g_{j_2}f_{i_3}g_{j_4}\dots f_{i_k}$ and $g_{j_k}f_{i_{k-1}}g_{j_{k-2}}f_{i_{k-3}}\dots g_{j_1}$ are mutual inverses of $C$.

In the interest of having an undeleted answer, here is a small result. Let $x, y$ be objects and $f, g : x \to y$ and $u, v : y \to x$ be morphisms in $C$, and let

$$F = af + bg, G = cu + dv$$

be two morphisms in $RC$, where $a, b, c, d \in R$. If $FG = \text{id}_y, GF = \text{id}_x$, then WLOG $fu = \text{id}_y$ and also some term in $GF$ must equal $\text{id}_x$. If we want $x, y$ to be non-isomorphic, then $f$ cannot have a left inverse and $u$ cannot have a right inverse, so it must be the case that $vg = \text{id}_x$ and moreover no other composition of morphisms except $fu$ or $vg$ can be an identity.

It follows that $ac = bd = 1$, hence $a, b, c, d$ are all units, so none of the four terms in $FG$ or in $GF$ vanish. Thus the only way for all of the non-identity terms to cancel is if $gu = fv = gv$ and $ug = vf = vg$. But this implies

$$gug = fvg = f = gvg = g$$

and symmetrically $u = v$, so in fact $x, y$ must be isomorphic in $C$. Next on the list is linear combinations of three morphisms...