Yesterday I asked a very similar question about an exercise of Gelfand's book "Methods of Homological Algebra". In the comments it was pointed out that there was an easier version of that exercise but I couldn't solve it.
Given $f: K^{\bullet} \to L^{\bullet}$ given by the two upper rows of:
I have to show that this map is not zero in $D(\mathcal{A})$ ie there is no quasi isomorphism $s$ such that $sf$ is homotopic to $0$.
I know that the lowest row is acyclic in all degrees except 0 since $s$ is a quasi isomorphism. Also if we denote $d^{A}$ the differential of the last complex it's clear that $s_1$ factors through $Ker(d^{A}_0)$.
If we denote $t$ the homotopy map such that $s_1=d^{A}_{-1} t_1 + t_0 2$ and $0=d^{A}_0 t_0$ we can also see that $t_0$ factors through the kernel. I don't know how to get more information to find a contradiction.
Any tips to finish this? Also I don't know if this approach will be useful should I consider some mapping cones ?
Since $s$ is a quasi-isomorphism, you know that $$\bar s:\Bbb Z \to H^0(A)$$ is an isomorphism. Let $x\in A_0$ be $s_1(1)$. Then $\bar s(1)=x+Im(d^{-1})$ generates $H^0(A)\cong \Bbb Z$.
Let $y=t_1(1)$ and $z=t_0(1)$, then $d^{-1}y+2z=x$. As a consequence, you have that $x=2z$ mod $Im(d^{-1})$, i.e., $x=2z$ in $H^0(A)$. But that's a contradiction since $x$ is a generator of $H^0(A)\cong \Bbb Z$ (you can't divide 1 by 2 in $\Bbb Z$).
Therefore there is non such homotopy $t$.