Universal property of a quotient by a Serre subcategory I'm trying to deduce that the localization of a abelian category $\mathsf{A}$ with respect to the multiplicative system given by a Serre subcategory $\mathsf{B}$ satisfies the classical universal property of quotients of abelian categories:
If $\mathsf{Q_B}\colon \mathsf{A}\to \mathsf{A/B}$ is a localization functor, then it is exact, and for every exact functor $\mathsf{F}\colon \mathsf{A}\to \mathsf{C}$ there is a unique exact functor $\overline{\mathsf{F}}\colon\mathsf{A/B}\to\mathsf{C}$ for which we have $\overline{\mathsf{F}}\circ \mathsf{Q_B} = \mathsf{F}$
My main problem is seeing why the induced functor $\overline{\mathsf{F}}$ is exact. It would be easy if $\mathsf{Q_B}$ reflected exact sequences, but I doubt this is true. Let $\mathsf{S_B} = \{ f \in \mathsf{Mor(A)} \mid \mathrm{ker}(f), \mathrm{coker}(f) \in \mathsf{B} \}$. Then every morphism $\phi\colon A\to B$ in $\mathsf{A/B}$ may be represented as a right roof $\mathsf{Q_B}(f)\circ \mathsf{Q_B}(s)^{-1}$. It can be shown that $\mathrm{im}(\phi) = \mathsf{Q_B}(\mathrm{im}(f))$ and $\mathrm{ker}(\phi) = \mathsf{Q_B}(s)\circ \mathsf{Q_B}(\mathrm{ker}(f))$. Therefore, if $A \xrightarrow{\phi} B \xrightarrow{\psi} C$ is exact in $\mathsf{A/B}$ where $\phi = \mathsf{Q_B}(f)\circ \mathsf{Q_B}(s)^{-1}$ and $\psi = \mathsf{Q_B}(g)\circ \mathsf{Q_B}(t)^{-1}$, then $\mathrm{Q_B}(\mathrm{im}(f))$ and $\mathsf{Q_B}(t)^{-1}\circ \mathsf{Q_B}(\mathrm{ker}(g))$ are connected by an isomorphism. However, this isomorphism is an isomorphism in $\mathrm{A/B}$, not in $\mathrm{A}$.
Further than that, I don't have any ideas.
 A: You are of course correct that $\overline F$ does not reflect exact sequences, for instance take $s\in S_B$ which is not a monomorphism then $0\to x'\overset{s}\to x \to y\to z\to 0$ will not be an exact sequence, but if $0\to x\to y\to z\to 0$ is exact, then the former will be an exact sequence in the quotient.
The point is that any exact sequence in the quotient can be lifted (up to isomorphism) to an exact sequence in $A$.
A proof of that can for instance go as follows: first prove that any exact sequence in $A/B$ is the image of a sequence in $A$ (not necessarily exact). This can be proved using the representation of morphisms in $A/B$ as $fs^{-1}$ and $s^{-1}f$.
Second, use the fact that the quotient functor $A\to A/B$ is exact to modify the sequence in $A$ to be an exact sequence with the same image. Namely, suppose $0\to x\to y\to z\to 0$ is a sequence in $A$ whose image is exact in $A/B$, then define $x' := \ker(y\to z)$, there is a canonical isomorphism in $A/B$ (not necessarily in $A$, it's not even clear that the composite $x\to y\to z$ is $0$ in $A$) between this sequence and $0\to x'\to y\to z\to 0$, and then similarly replace $z$ with $z':=\mathrm{coker}(x'\to y)$ to get an exact sequence $0\to x'\to y\to z'\to 0$ in $A$ whose image is your original one.
