We will use the description of the category $\mathcal{A}/\mathcal{B}$ in the proof of Homology, Lemma \ref{homology-lemma-serre-subcategory-is-kernel}. Let $(X^\bullet, d^\bullet)$ be a complex of $\mathcal{A}/\mathcal{B}$. This means that $X^i$ is an object of $\mathcal{A}$ and $d^i : X^i \to X^{i + 1}$ is a morphism in $\mathcal{A}/\mathcal{B}$ such that $d^i \circ d^{i - 1} = 0$ in $\mathcal{A}/\mathcal{B}$. \medskip\noindent For $i \geq 0$ we may write $d^i = (s^i, f^i)$ where $s^i : Y^i \to X^i$ is a morphism of $\mathcal{A}$ whose kernel and cokernel are in $\mathcal{B}$ (equivalently $s^i$ becomes an isomorphism in the quotient category) and $f^i : Y^i \to X^{i + 1}$ is a morphism of $\mathcal{A}$. By induction we will construct a commutative diagram $$ \xymatrix{ & (X')^1 \ar@{..>}[r] & (X')^2 \ar@{..>}[r] & \ldots \\ X^0 \ar@{..>}[ru] & X^1 \ar@{..>}[u] & X^2 \ar@{..>}[u] & \ldots \\ Y^0 \ar[u]_{s^0} \ar[ru]_{f^0} & Y^1 \ar[u]_{s^1} \ar[ru]_{f^1} & Y^2 \ar[u]_{s^2} \ar[ru]_{f^2} & \ldots } $$ where the vertical arrows $X^i \to (X')^i$ become isomorphisms in the quotient category. Namely, we first let $(X')^1 = \Coker(Y^0 \to X^0 \oplus X^1)$ (or rather the pushout of the diagram with arrows $s^0$ and $f^0$) which gives the first commutative diagram. Next, we take $(X')^2 = \Coker(Y^1 \to (X')^1 \oplus X^2)$. And so on. Setting additionally $(X')^n = X^n$ for $n \leq 0$ we see that the map $(X^\bullet, d^\bullet) \to ((X')^\bullet, (d')^\bullet)$ is an isomorphism of complexes in $\mathcal{A}/\mathcal{B}$. Hence we may assume $d^n : X^n \to X^{n + 1}$ is given by a map $X^n \to X^{n + 1}$ in $\mathcal{A}$ for $n \geq 0$. \medskip\noindent Dually, for $i < 0$ we may write $d^i = (g^i, t^{i + 1})$ where $t^{i + 1} : X^{i + 1} \to Z^{i + 1}$ is an isomorphism in the quotient category and $g^i : X^i \to Z^{i + 1}$ is a morphism. By induction we will construct a commutative diagram $$ \xymatrix{ \ldots & Z^{-2} & Z^{-1} & Z^0 \\ \ldots & X^{-2} \ar[u]_{t_{-2}} \ar[ru]_{g_{-2}} & X^{-1} \ar[u]_{t_{-1}} \ar[ru]_{g_{-1}} & X^0 \ar[u]_{t^0} \\ \ldots & (X')^{-2} \ar@{..>}[u] \ar@{..>}[r] & (X')^{-1} \ar@{..>}[u] \ar@{..>}[ru] } $$ where the vertical arrows $(X')^i \to X^i$ become isomorphisms in the quotient category. Namely, we take $(X')^{-1} = X^{-1} \times_{Z^0} X^0$. Then we take $(X')^{-2} = X^{-2} \times_{Z^{-1}} (X')^{-1}$. And so on. Setting additionally $(X')^n = X^n$ for $n \geq 0$ we see that the map $((X')^\bullet, (d')^\bullet) \to (X^\bullet, d^\bullet)$ is an isomorphism of complexes in $\mathcal{A}/\mathcal{B}$. Hence we may assume $d^n : X^n \to X^{n + 1}$ is given by a map $d^n : X^n \to X^{n + 1}$ in $\mathcal{A}$ for all $n \in \mathbf{Z}$. \medskip\noindent In this case we know the compositions $d^n \circ d^{n - 1}$ are zero in $\mathcal{A}/\mathcal{B}$. If for $n > 0$ we replace $X^n$ by $$ (X')^n = X^n/\sum\nolimits_{0 < k \leq n} \Im(\Im(X^{k - 2} \to X^k) \to X^n) $$ then the compositions $d^n \circ d^{n - 1}$ are zero for $n \geq 0$. (Similarly to the second paragraph above we obtain an isomorphism of complexes $(X^\bullet, d^\bullet) \to ((X')^\bullet, (d')^\bullet)$.) Finally, for $n < 0$ we replace $X^n$ by $$ (X')^n = \bigcap\nolimits_{n \leq k < 0} (X^n \to X^k)^{-1}\Ker(X^k \to X^{k + 2}) $$ and we argue in the same manner to get a complex in $\mathcal{A}$ whose image in $\mathcal{A}/\mathcal{B}$ is isomorphic to the given one.