Let $D(A)$ be the derived category of an abelian category $A$ with the homotopy chain complex category $K(A)$. I want to characterize when morphisms in $K(A)$ become isomorphisms in $D(A)$. We can write morphisms in $D(A)$ as $gf^{-1}$, where $g,f$ are morphisms in $K(A)$ and $f$ is a quasi-isomorphism. The claim is that $gf^{-1}$ is an isomorphism if and only there are two morphisms $l,r$ such that both $lg$ and $gr$ are quasi-isomorphisms in $K(A).$
This is basically Lemma 5.3.1 here, written in the language of derived categories instead of the more general language of Verdier quotients. The 'if' direction is clear. The existence of $r$ is also clear: if $gf^{-1}$ is invertible, then $g$ is also invertible in the derived category, so there is some right inverse $rs^{-1}$ and then $gr$ is a quasi-isomorphism. However, I think there is something wrong with the existence of $l$.
Let's try doing the same thing in the other direction. We have a left inverse $lh^{-1}$, then $lh^{-1}g$ is the identity in the derived category, so there is some equivalent morphism of the form $l\tilde{g}\tilde{h}^{-1}$ in which $l\tilde{g}$ is a quasi-isomorphism. This is obviously not what we wanted, but I also don't see how we can do better than that. If $g:X\to Y$, whenever I try writing morphisms as roofs I find that there are simply no morphisms out of $Y$ to compose with.
At the same time I expected the statement to be true. I think if we describe the same derived category using co-spans, the existence of $l$ will become trivial, and we will run into the same problem trying to find $r.$ Basically, I think that the statement is only true because we can describe $D(A)$ in both ways, using either the calculus of left fractions or right fractions. Am I missing something? Or is there something wrong with this statement?