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I have for any $n$ some distinguished triangles $$X_n->Y_n->Z_n$$ $$X'_n->Y'_n->Z'_n$$ The families $(X_n), (Y_n), (X'_n), (Y'_n)$ are inductive systems, $(X_n)$ is quasi-isomorphic to $(X'_n)$, $(Y_n)$ is quasi-isomorphic to $(Y'_n)$ and these quasi-isomorhisms are compatible with the maps $(u_n):(X_n)->(Y_n)$ and $(v_n):(X'_n)->(Y'_n)$.

Can I deduce a quasi-isomorphism of inductive systems between $(Z_n)$ and $(Z'_n)$? If not any other conditions on $(u_n)$ and $(v_n)$ that would allow to get the same conclusion?

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In what category are you working? Is this the derived category (or chain homotopy category) of an abelian category, or an arbitrary triangulated category? If it's the former, then you can use the cone construction to choose distinguished maps on the $Z$'s in a functorial way, so you can make $(Z_n)$ and $(Z_n')$ into inductive systems with a map between them. If it's the later, I'm not sure if there necessarily exists a functorial cone construction. – Anton Geraschenko Feb 6 '11 at 17:02
@Anton: I had the same qualms. Yes, if you're actually given an inductive system of chain maps, then you get an inductive system on the strict triangles (and if you can reduce to this situation then the answer to the question is trivially yes). But already if you pass to homotopy classes (let alone to the derived category), you're in deep trouble because the cone is no longer functorial except in silly situations and I don't see how it should be possible to lift an inductive system of homotopy classes to an inductive system of actual chain maps. – Theo Buehler Feb 6 '11 at 17:26
Question to Anton: Even if I am in your former case and can replace Z_n and Z'_n by cones is it clear that I can choose the maps between cones to satisfy Z_n->Z_{n+1}->Z'_{n+1} = Z_n->Z'_n->Z'_{n+1} (I do need this if I want to get a map of inductive system, right?) – user12770 Feb 7 '11 at 0:13
I hadn't carefully gone through the details. It sounds like I was just wrong. A homotopy between two chain maps induces a chain isomorphism between their cones. Moreover, if you have a morphism of morphisms in the homotopy category (i.e. a homotopy commutative square), I think the homotopy making the square commutative induces a morphism of cones. To get an induced functor in the homotopy category, you'd need to show that different homotopies induce homotopic maps between the cones. @Theo: it sounds like you're claiming this is actually false. Is there a good way to see that? – Anton Geraschenko Feb 7 '11 at 18:07
@Anton: What you say is correct. Here's how I think about it: The composition of consecutive morphisms in a distinguished triangle is zero and TR3 essentially tells us that the cone is a weak cokernel. The morphism on the cone is unique if $\text{Hom}_{\mathbf{K}}^{-1}(X,Z') = 0$ because $\text{Hom}_{\mathbf{K}}({-},Z')$ is homological, see Bernstein-Beilinson-Deligne, 1.1.9. Explicit counterexamples are a little messy to come up with. There might be one in Neeman's "Some new axioms for triangulated categories". – Theo Buehler Feb 9 '11 at 11:58

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