Multiplicative system of a ring and of a category If A is any category, a class of morphisms $S$ in A is said to be a multiplicative system whether
$(a)$ it's closed by composition, that is: $id_X$ is in $S$ for every $X$ in A and whenever $f$ and $g$ are morphisms in A such that the composition $gf$ makes sense, then $gf$ is in $S$; $(b)$ any diagram of the form $X\overset{f}\longrightarrow Y \overset{s}\longleftarrow Z$ with $s$ in $S$ can be completed as
$\require{AMScd}$
\begin{CD}
W @>g>> Z\\
@VtVV @VVsV\\
X @>>f> Y
\end{CD}
with $t$ in $S$. The same also with all the arrows reversed. Eventually $(c)$ for a pair of morphisms $f,g:X\to Y$ there exists $s$ in $S$ with $sf=sg$ if and only if there exists $t$ in $S$ with $ft=gt$.
My question is: does this definition coincide with the notion of multiplicatively closed set for any ring $R$ if we look at $R$ as an Ab-category with just one object? Certainly condition $(a)$ provides exactly what we desire for a multiplicatively closed set (that is a subset $S\subseteq R$ such that $1\in S$ and $x,y\in S\Rightarrow xy,yx\in S$), and if $R$ is commutative, $(b)$ and $(c)$ become obvious, but in the case of a non-commutative ring I can't find a proof of these conditions.
Could anyone provide a proof or a counterexample? If a counterexample is the answer, is there any profound reason why it happens to works only in the commutative case, or it's the notion of multiplicative system to be designed just to generalize these cases?
 A: Yes, it coincides, but rather trivially (in the commutative case).
See your (commutative unital) ring $R$ as a category as follows. The $R$-module action of $R$ on itself induces a morphism $\iota: R \to \mathrm{Hom}_{\mathbb Z}(R,R)$, so we can consider the category with one object (namely $R$) and the set of morphisms is $\iota(R)$. The fact that this forms an $\mathbf{Ab}$-category is part of the axioms of a ring. You need the ring to be unital for the identity morphism to be present, and commutativity gives you the other axioms. For instance, if you are given
$\require{AMScd}$
\begin{CD}
R @>f>> R @<s<< R
\end{CD}
you are basically given two elements of the original ring $R$. The diagram can easily be completed by assuming that $R$ is commutative since $sf = fs$ leads to the commutative diagram
$\require{AMScd}$
\begin{CD}
R @>f>> R\\
@VsVV @VVsV\\
R @>>f> R
\end{CD}
Statement (c) is proven similarly by taking $t=s$. I don't know about localizing non-commutative rings at subsets $S$ in general, but I would bet that if these ideas made sense, then the localzation $S^{-1}R$ would exist when $R$ is non-commutative in the specific case where those categorical axioms are satisfied, but not in general. I read this to know a bit about non-commutative localization, and it doesn't feel as inspiring as the commutative counterpart.
Hope that helps,
