The construction of the localization of a category

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).

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After some consideration, it appears to me that inverting a small set $\mathcal{M}$ of morphisms in a locally small category $\mathcal{C}$ always gives another locally small category $\mathcal{D}$. Assume $\mathcal{M}$ is closed under composition – this can be done by induction and we still get a small set.

Firstly, notice that any morphism $X \to Y$ in $\mathcal{D}$ can be expressed as a finite zigzag
$$X \rightarrow X_1 \leftarrow X_2 \rightarrow X_3 \leftarrow \cdots \rightarrow Y$$
of morphisms in $\mathcal{C}$, where the leftward arrows are taken from the set $\mathcal{M}$. In particular, the intermediate objects $X_1, X_2, X_3, \ldots$ must be drawn from a small set, namely $\{ \operatorname{dom} f : f \in \mathcal{M} \} \cup \{ \operatorname{codom} f : f \in \mathcal{M} \}$. (Use the axiom of replacement!) Thus we can construct a surjection from a small set to $\mathcal{D}(X, Y)$, and it follows that $\mathcal{D}(X, Y)$ is a small set. However, in practice it’s not so easy to describe exactly which zigzags get identified…

In applications $\mathcal{M}$ is usually not a small set, so we are back to square one. It’s not clear to me whether there is an explicit example of a locally small category whose localisation is not locally small – perhaps this is worth asking as a separate question.