Calculus of fractions and Serre subcategories Let $\mathcal{A}$ be an abelian category, and $\mathcal{B}\subseteq\mathcal{A}$ a Serre subcategory. I want to show that the set of arrows in $\mathcal{A}$

$$S = \{ f \in \text{Arrows}(\mathcal{A}) \mid \mathop{\mathrm{Ker}}(f), \mathop{\mathrm{Coker}}(f) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{B}) \}$$
  admits the calculus of a fractions. 

I have found a proof of it in the Stacks project, but the definition of a Serre subcategory given there is different from the one I know, so that I cannot apply the same proof. 
 A: Let $\mathcal{A}$ be an abelian category and $\mathcal{B}$ a (nonempty) full subcategory of $\mathcal{A}$. As Max pointed out, it suffices to show that the following are equivalent:
i) for any exact sequence of the type
$$ 0\to M\to M'\to M''\to 0$$
$M'$ is in $\mathcal{B}$ if and only if $M$ and $M''$ are in $\mathcal{B}$.
ii) for any exact sequence of the type
$$M\to M'\to M''$$
with $M$ and $M''$ in $\mathcal{B}$, then also $M'$ is in $\mathcal{B}$.
The implication ii) $\Rightarrow$ i) is clear. To prove that i) $\Rightarrow$ ii), we have to take an exact sequence of the form $M\xrightarrow{f} M'\xrightarrow{g} M''$, with $M, M'' \in \mathcal{B}$, and show that $M'$ is in $\mathcal{B}$, too. Consider the short exact sequence
$$0\to \ker g \to M' \to \text{im}\ g\to 0$$
and notice that 


*

*$\ker g\in \mathcal{B}$, by applying i) to the s.e.s. $0 \to \ker f \to M \to \text{im}\ f \to 0$ and using the isomorphism $\text{im}\ f \cong \ker g$,

*$\text{im}\ g \in \mathcal{B}$, with the same argument for the s.e.s. $0\to \text{im}\ g = \ker \text{coker}\ g\to M'' \to \text{im}(\text{coker}\ g)\to 0$, 
therefore by i) we have $M'\in\mathcal{B}$.
