Localisation of inclusion functors Let $\mathcal C$ be a category and suppose $\cal B \subseteq C$ is a full subcategory. Let $i \colon \mathcal B \longrightarrow \cal C$ denote the inclusion functor. Suppose that $S \subseteq \operatorname{Mor}\mathcal C$ is a class of morphisms in $\mathcal C$. Then we get a functor 
$$\tilde i \colon \mathcal B[(S \cap \operatorname{Mor} \mathcal B) ^{-1}] \longrightarrow \mathcal C [S^{-1}]$$
 If $S$ satifies the Ore conditions, and if for every morphism $s \colon M \longrightarrow N$ in $S$ with $M \in \mathcal B$ there exists a morphism $u \colon N \longrightarrow P$ such that $us \in S$ and $P \in \mathcal B$, then this functor $\tilde i$ is fully faithful.
Are there more general results that say when $\tilde i$ is fully faithful?
 A: Let $\mathscr{C}$  a category  and let $\Sigma \subset   \mathscr{C}$ be a wide subcategory  (i.e. closed under composition and containing identites).
There exists (generally in a more large sets universe) the category of fractions $P: \mathscr{C} \to  \mathscr{C}(\Sigma )$, and $P$ is the identity map on objects. Given a morphism $\hat{f}\in  \mathscr{C}(\Sigma )(A, B)$ we say that a sequence  $A  \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_2\xleftarrow{s_2} Y_2 \ldots \xleftarrow{s_n} Y_n \xrightarrow{f_n} B$ in $\mathscr{C}$ with $s_i \in  \Sigma \ 1 \leq  i\leq n$, represent $\hat{f} $ if $P(f_n)\circ P(s_n^{-1} )\circ \ldots P(s_1^{-1} )\circ P(f_1)= \hat{f}  $. From the construction of $\mathscr{C}(\Sigma )$ any $\hat{f}$ is represented by some sequences.
1) Let $\Sigma$ with the right calculus of fraction.
Then any $\hat{f}$ has a short representation as a span like $A\xrightarrow{s} X \xleftarrow{f} B$, $s\in \Sigma$, we call it a $\Sigma$-span and indicate it as $(A, s, X, f, B)$ or simply as $(s, X, f)$.
We ask when two $\Sigma$-spans are equivalent i.e. represent the some morphisms in $\mathscr{C}(\Sigma )$.
The answere is that two $\Sigma$-spans $(A, s, X, f, B)$, $(A, t, Y, g, B)$ are equivalent iff exist a commutative diagram in  $\mathscr{C}$:
$$
\begin{array}{ccccccc}
 E& \xrightarrow{c} & X & \xrightarrow{f} & B  \\
\parallel&& \downarrow     s & &\parallel \\
E &\xrightarrow{u}& A &   & B \\
\parallel && \uparrow t     & & \parallel\\
E& \xrightarrow[d]{} & Y & \xrightarrow{g} & B  \\
\end{array}$$
with $u\in \Sigma$ and $c, d \in \mathscr{C}$. (the proof is very tedious, in Gabriel Zisman there isn't a true proof, omit  not obvious details, and in H. Shubert there is only a incomplete proof)
In other words $\mathcal{C}(\Sigma )(A, B) \cong  \varinjlim_{(X, s)\in \mathscr{C} _\Sigma (A)^{op}} (X, B) = \varinjlim\ (\ \mathscr{C}_\Sigma(A)^{op} \xrightarrow{\pi } \mathscr{C}^{op} \xrightarrow{[-, B]} Set\ )  $ where  $\mathscr{C}_\Sigma (A)$ is the full subcategory of $A \downarrow \mathscr{C}$ with objects the arrow that belong to $\Sigma $ , observe that $\mathscr{C}_\Sigma(A)^{op}$ is  filtered.
Now consider the natural map $\mu : \mathscr{B}(\Sigma \cap  \mathscr{B})(A, B) \to  \mathscr{C}(\Sigma )(A, B) $ this is surjective iff for each $\Sigma $-span $(s, X, f)$ there is a equivalent $\Sigma $-span $(t, Y, g)$ with $Y \in \mathscr{B}$ (i.e. a diagram as above). And $\mu $ is injective iff given a diagram as above with $X, Y\in \mathscr{B}$ then exist a similar diagram with $E\in \mathscr{B}$. Of course $\mu $ is bijective if $\mathscr{B}_\Sigma (A)^{op} \to  \mathscr{C}_\Sigma (A)^{op}$ is a final functor.
2) $\Sigma $ without  "calculus of right fractions" hypothesis.
suppose that for  $A\in \mathscr{B}$, $Y \in  \mathscr{C}$ each span
$A \xleftarrow{s} Y \xrightarrow{f} X$ is connected to a span like $ A \xleftarrow{s'} B \xrightarrow{f'} X $ with $B \in  \mathscr{B}$, and
each cospan
$A \xrightarrow{f}  X \xleftarrow{s} Y $ is connected to a cospan like $ A \xrightarrow{f'} B \xleftarrow{s'} Y $ with $B \in  \mathscr{B}$.
I claim that the map $\mu : \mathscr{B}(\Sigma \cap  \mathscr{B})(A, B) \to  \mathscr{C}(\Sigma )(A, B) $  is surjective.
Given $\hat{f}: A \to  B$, chose  a representing sequences $A  \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_1\xleftarrow{s_2} Y_2 \ldots \xleftarrow{s_n} Y_n \xrightarrow{f_n} B$,
Now the span $A \xleftarrow{s_1} Y_1 \xrightarrow{f_1} X_1$ is connected by a span like $A \xleftarrow{s'_1} B_1 \xrightarrow{f'_1} X_1$ with $B_1 \in  \mathscr{B}$ and we replace this second to first, and consider the cospan $B_1 \xrightarrow{f'_1} X_1 \xleftarrow{s_2}$ this is connected to a cospan  like $B_1 \xrightarrow{f''_1} A_1 \xleftarrow{Y_2}$ with $A_1 \in  B$, and so on.. at the end we have a sequence $A  \xleftarrow{s'_1} B_1 \xrightarrow{f''_1} A_1\xleftarrow{s'_2} B_2 \ldots \xleftarrow{s'_n} B_n \xrightarrow{f''_n} B$ in $\mathscr{B}$ such that mapped by $P$ in $\mathscr{C}( \Sigma )$ (and inverting all the $s'_i$ $1 \leq i\leq n$) obtain  $\hat{f} $.
