For those who care, here's the proof that $j_{M^\bullet}$ is a qis (as it is hinted the last sentence of the proof).

> **Definition.** Extending the terminology introduced in Derived Categories, Remark $\ref{09WA}$, if $\mathcal{A}$ is an abelian category, a triangle $A\to B\to C\to A[1]$ in $K(\mathcal{A})$ or in $D(\mathcal{A})$ is said to be *$H^*$-special* if its induced long sequence in cohomology is exact.

We know that all distinguished triangles in $K(\mathcal{A})$ and in $D(\mathcal{A})$ are $H^*$-special. The converse isn't true in general: Let $\varepsilon_i\in\{1,-1\}$ be any signs satisfying $\varepsilon_1\varepsilon_2\varepsilon_3=-1$. If $A\xrightarrow{a}B\xrightarrow{b}C\xrightarrow{c} A[1]$ is a d.t. in $K(\mathcal{A})$ or in $D(\mathcal{A})$, then the anti-distinguished triangle $A\xrightarrow{\varepsilon_1 a}B\xrightarrow{\varepsilon_2 b}C\xrightarrow{\varepsilon_3 c} A[1]$ is again $H^*$-special but not necessarily distinguished. Note that if $(f,g,h)$ is a morphism of $H^*$-special triangles and at least two among $f,g,h$ are quasi-isomorphisms, then so is the third.

Now, consider the following diagram in $D(\mathcal{A})$:
$$\require{AMScd} \begin{CD} J^\bullet(M^\bullet)@>>> C(u_{M^\bullet})@>-p>>M^\bullet[1]@>u_{M^\bullet}>>J^\bullet(M^\bullet)[1]\\ @|@VVqV@|@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash{-pq^{-1}}>>M^\bullet[1]@>\smash{u_{M^\bullet}}>>J^\bullet(M^\bullet)[1]\\ @|@|@VVj_{M^\bullet}[1]V@|\\ J^\bullet(M^\bullet)@>\smash{v_{M^\bullet}}>>Q^\bullet(M^\bullet)@>\smash i>>C(v_{M^\bullet})@>\smash{(0,1)}>>J^\bullet(M^\bullet)[1] \end{CD}$$
where $j_{M^\bullet}=\begin{pmatrix}0\\u_{M^\bullet}\end{pmatrix}$, $p=(0,1)$ and $q=(v_{M^\bullet},0)$ is a quasi-isomorphism as it is argued in Derived Categories, Section $\ref{014Z}$, before Lemma $\ref{0152}$. In the diagram, the top row is a distinguished triangle. Since the three upper squares commute and $q$ is an iso in $D(\mathcal{A})$, the middle row is a distinguished triangle too (in $D(\mathcal{A})$). On the other hand, the bottom row is anti-distinguished, so it is $H^*$-special. Thus, to see that $j_{M^\bullet}[1]$ is a qis, it suffices to see that the three lower squares commute. Clearly the left and right lower squares commute. Commutativity of the middle square amounts to $i\circ q=-j_{M^\bullet}[1]\circ p$. It suffices to verify this equality in $K(\mathcal{A})$. This is the content of the following

> **Lemma.** Let $A\xrightarrow{a} B\xrightarrow{b} C$ be morphisms in some additive category. Then the morphisms $\begin{pmatrix}b&0\\0&0\end{pmatrix},\begin{pmatrix}0&0\\0&-a[1]\end{pmatrix}:C(a)\rightrightarrows C(b)$ are chain homotopic.

*Proof.* A homotopy is given by $$h^n=\begin{pmatrix}0&0\\1_{B^n}&0\end{pmatrix}:C(a)^n=B^n\oplus A^{n+1}\to C^{n-1}\oplus B^n=C(b)^{n-1}.\quad\square$$