I was reading the construction of the localization of a category in the book "Methods of homological algebra" of Manin and Gelfand.
Let me remind you the definition of the localization of a category: Let $B$ be an arbitrary category and $S$ a set of morphisms in $B$, the localization is a category $B[S^{-1}]$ with a functor $Q:B\rightarrow B[S^{-1}]$ such that $Q(s)$ is an isomorphism for every $s\in S$, and if another functor $F:B\rightarrow D$ has this property then there exists a functor $G:B[S^{-1}]\rightarrow D$ such that $F=G\circ Q$.
Before asking my question let me remind you the construction of the localization:
Let $B$ be an arbitrary category and "S" an arbitrary set of morphisms in $B$, we want to construct $B[S^{-1}]$. Set $\mathrm{Ob}\;B[S^{-1}]=\mathrm{Ob}\;B$, and define $Q$ to be the identity on objects. I want to construct the morphisms of $B[S^{-1}]$, to do that we proceed in several steps:
a) introduce variables $x_s$, one for every morphism $s\in S$.
b) Now construct an oriented graph $\Gamma$ as follows:
vert $\Gamma$=Ob $B$
Edges of $\Gamma$=$\{$morphisms in $B\}\cup\{x_s:s\in S\}$
if $f$ is a morphism $X\rightarrow Y$ then it correspondes to an edge $X\rightarrow Y$
if $s\in S$ then $x_s$ is an edge $Y\rightarrow X$.
A path in this graph is what you expect it to be.
Now lets define an equivalence relation among paths with the same beginning and same ending. We say that two paths are equivalent if they can be joined by a chain of these two types of operations:
1) two consecutive arrows can be replaced with their composition
2) the path $sx_s$ is equivalent to $id$ and the same for $x_ss$.
So a morphism is an equivalence class of paths with the common beginning and common end.
Now if you are interested you can easily continue defining $Q$ and proving that this category has the property that we want.
My question is this: are we sure that this is a category? Because the class of morphisms can be a proper class, it's not obvious to me that is a set. I was told that there is a way to fix this (or at least to bypass the problem); do you know a way to fix this issue?
(I know that for example if the set $S$ has good properties, i.e. it's a localizing system of morphisms, then there is a nicer construction, but I would like to know a general construction that works even if $S$ is not localizing).