I am the reading the paper Chen, Xiao-Wu, and Henning Krause. "Introduction to coherent sheaves on weighted projective lines." arXiv preprint arXiv:0911.4473 (2009).
Let $\mathscr{C}$ be a Serre subcategory of the abelian category $\mathscr{A}$ and $A/A',B'\in\mathscr{C}$ with $A'\subset A,B'\subset B$ subobjects. Furthermore, let $A/A'$ and $B'$ both in $\mathscr{C}$ constitute a directed set allowing to form a direct system of abelian groups $\text{Hom}_{\mathscr{A}}(A',B/B')$. The quotient category $\mathscr{A}/\mathscr{C}$ are defined by applying a functor taking the identity on the objects of $\mathscr{A}$ and mapping a morphism in
$$\text{Hom}_{\mathscr{A}}(A,B)$$
naturally to the colimit in this direct system of abelian groups $\text{Hom}_{\mathscr{A}}(A',B/B')$.
This can be done by sending a morphism $f$ to the equivalence class $\{\pi_{B'}\circ f\circ\iota_{A'}\}$, where $\pi_{B'}$ quotients by $B'$ and $\iota_{A'}$ is an embedding into $A$, under the appropriate equivalence relation deduced from the above direct system of abelian groups.
On page 5 of this paper it states that "The quotient functor sends a morphism in $\mathscr{A}$ to the zero morphism if and only if its image belongs to $\mathscr{C}$." I have problems with the implication part, the converse part is fine.
I tried to make a small drawing, trying to demonstrate the equivalence relation at work which make the diagram (neglecting the image object of $f$ in the diagram) commutative. In this diagram $C$ is a subobject of $A$ and $A'$ and $D\subseteq D'$ are subobjects of $B$. Furthermore $A/C,A/A'$ and $D',D$ are all in the Serre subcategory $\mathscr{C}$. We have to show that $\text{Im}(f)$ is an object in $\mathscr{C}$. A Serre subcategory is closed under taking subobjects, quotients and extensions.
Please apologise the small inconsistency in notation regarding the quotient morphism $\pi$ and the embeddings $\iota$. I replaced $\pi$ with $p$ and $\iota$ with $i$, using appropriate subscripts.
One approach was the following. I tried to write down a short exact sequence with $\text{Im}(f)$ as an extension of two object in $\mathscr{C}$, unfortunately not succesfully so far. One further idea was to include the object $A/C$ or $A/A'$ (as an object in the Serre category $\mathscr{C}$ in the short exact sequence.
Can you give me an indication how to tackle this question?
If $f:A\to B$ is zero in the quotient category, then there are subobjects $A'$ of $A$ and $B'$ of $B$ such that $A/A'$ and $B'$ are in $\mathscr{C}$ and such that the composition $$A'\hookrightarrow A\xrightarrow{f}B\twoheadrightarrow B/B'$$ is zero.
So $A'\hookrightarrow A\xrightarrow{f}B$ factors through $B'$ and $A\xrightarrow{f}B\twoheadrightarrow B/B'$ factors through $A/A'$, and we get a commutative diagram with exact rows: $\require{AMScd}$ \begin{CD} 0@>>>A'@>>>A@>>>A/A'@>>>0\\ @.@VVV@VVfV@VVV@.\\ 0@>>>B'@>>>B@>>>B/B'@>>>0 \end{CD}
Let $K=\ker{f}$, so $\operatorname{im}f\cong A/K$, and so we need to show that $A/K$ is in $\mathscr{C}$.
By commutativity of the left hand square, $A'/(A'\cap K)$, which is the image of the composition $A'\hookrightarrow A\xrightarrow{f}B$, is a quotient of a subobject of $B'$, which is in $\mathscr{C}$. So $A'/(A'\cap K)$ is in $\mathscr{C}$.
By the second isomorphism theorem, $A'/(A'\cap K)\cong (A'+K)/K$, so $(A'+K)/K$ is in $\mathscr{C}$.
$A/(A'+K)$ is a quotient of $A/A'$, which is in $\mathscr{C}$, so $A/(A'+K)$ is in $\mathscr{C}$.
Finally, $A/K$ is an extension of $A/(A'+K)$ and $(A'+K)/K$, so $A/K$ is in $\mathscr{C}$.