I set this problem in the framework of (pretriangulated) dg-categories; everything can probably be translated in the world of stable $(\infty,1)$-categories.

Let $\mathcal A$ be a pretriangulated dg-category. It is known that the homotopy category $H^0(\mathcal A)$ is triangulated, with exact triangles coming from the "pre-triangles" in $\mathcal A$. Assume we have a diagram in $\mathcal A$:

where the rows are pretriangles in $\mathcal A$, the vertical arrows are closed and of degree $0$, and everything is commutative in $H^0(\mathcal A)$. This diagram, in other words, induces a morphism of exact triangles in $H^0(\mathcal A)$.

Now, let $h : A \to B'$ be a degree $-1$ morphism such that \begin{equation} dh = vf - f'u, \end{equation} which exists by hypothesis. In the dg-category $\mathcal A$, there is a canonical (closed, degree $0$) morphism $C(u,v,h) : C(f) \to C(f')$ induced (functorially!) between the cones. This morphism also makes the diagram

commute in $H^0(\mathcal A)$. My question is the following: is it true or false, in general, that $[w] = [C(u,v,h)]$ as morphisms in $H^0(\mathcal A)$? Actually, there are some subtleties. Better said: can I choose $u': A \to A'$, $v': B \to B'$ closed of degree $0$ such that $[u]=[u']$, $[v]=[v']$, and $h' : A \to B'$ closed of degree $-1$ with $dh' = v'f - f'u'$, such that $[w] = [C(u',v',h')]$? I believe the answer of the above question is *false*, but perhaps it is true with some hypothesis on $\mathcal A$?

noin general, as you guess. Counterexamples are not that obvious though. If I have time (and can get a better connection where I am) I'll come back with one. – Fernando Muro Aug 5 at 15:55isinduced by some $u$ and $v$; the problem is that we are not assured that it is induced by thegiven$u$ and $v$. – Francesco Genovese Aug 5 at 20:05