# Serre quotient category

Let $\mathcal{A}$ be an abelian category. A Serre subcategory of $\mathcal{A}$ is a nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an exact sequence $$A \longrightarrow B\longrightarrow C$$ with $A$ and $C$ in $\mathcal{C}$, then also $B$ lies in $\mathcal{C}$.

(See the following tag of the Stacks Project: http://stacks.math.columbia.edu/tag/02MN)

Whenever we have a Serre subcategory $\mathcal{C}$ of an abelian category $\mathcal{A}$, we may form a "quotient category" $\mathcal{A}/ \mathcal{C}$. As far as I understand, there are different equivalent ways of constructing this quotient. Namely, one can look at it as a Gabriel-Zisman localisation of $\mathcal{A}$ at $$\{f\text{ morphism in }\mathcal{A}: \operatorname{Ker}f \text{ and }\operatorname{Coker}f \text{ are objects in }\mathcal{C}\}.$$ Indeed, this class is supposed to be a left and right multiplicative system, but I am having some trouble in checking the Ore condition.

Indeed, I can understand everything in the proof of the Lemma 12.9.6 in the link pasted above, except the following assertion (in the third paragraph of the proof): "then $\operatorname{Ker}t\longrightarrow \operatorname{Ker}s$ is surjective".

This must be very easy, but I am not seeing why it should be true.

It's easy enough using elements. First, $s(b)$ is the class of $(0,b)$ in $(C \oplus B)/\mathrm{Im}(t,-g)$. So given any $b \in \mathrm{Ker}(s)$, we must have $(0,b) \in \mathrm{Im}(t,-g)$, i.e., for some $a \in A$ we have $t(a) = 0$ and $-g(a)=b$. Then $-a \in \mathrm{Ker}(t)$ maps to $b \in \mathrm{Ker}(s)$.