Given a category $C$ and a commutative ring $R$, denote by $RC$ the $R$-linearization: this is the category enriched over $R$-modules which has the same objects as $C$, but the morphism module between two objects $x$ and $y$ is the free $R$-module on $\operatorname{Hom}\_C(x,y)$. Thus in $RC$ we allow arbitrary $R$-linear combinations of morphisms from the original category $C$.


Question: if two objects in $x$, $y \in C$ are isomorphic in $RC$, are they already isomorphic in $C$?


I do not know the answer to this question for *any* nontrivial ring $R$, but I'm particularly interested in $R=\mathbb{Z}$ and $R=\mathbb{Z}/2\mathbb{Z}$.


What's obviously not true is that every isomorphism in $RC$ comes from an isomorphism in $C$ (take $-id\_x$). (Thus the word "isomorphism" in the title refers to a relation on objects rather than to a property of morphisms.)


Of course, it is enough to consider categories $C$ with two objects $x$, $y$, but we cannot assume that $C$ is finite.


It's fairly elementary to see that if $x$ and $y$ are isomorphic in $RC$ then in $C$, $x$ is a retract of $y$ and vice versa, but the latter does in general not imply that $x \cong y$.


A more catchy way of phrasing this problem is: can we always classify objects in a category up to isomorphism by means of functors taking values in $R$-linear categories? (The inclusion $C \to RC$ is the universal such functor.)


*Edit*: A lot of people have posted an "answer" that wasn't, and deleted it, so here's something that will not work, to save others going down the same road. I said that we cannot assume that the category is finite; in fact, it must be infinite. Here is an elementary argument:


Since $x$ and $y$ are mutual retracts, there are maps $f,\;f'\colon x \to y$ and $g,\;g'\colon y \to x$ with $fg=\operatorname{id}$ and $g'f'=id$. Consider the powers of $fg' \in \operatorname{End}(y)$. If $\operatorname{End}(y)$ is finite then $(fg')^n = (fg')^m$ for some $m \neq n$; since $fg'$ has a right inverse (viz, $f'g$), we must have that $(fg')^n=\operatorname{id}$ for some $n>0$. So we see that $g'$ has not only a right inverse ($f'$) but also a left inverse: $(fg')^{n-1}f$. So they are the same and $g'$ is already an isomorphism. 



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 $C$.


