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If I have two exact triangles $X \to Y \to Z \to X[1]$ and $X' \to Y' \to Z' \to X'[1]$ in a triangulated category, and I have morphisms $X \to X'$, $Y \to Y'$ which 'commute' (i.e., such that $X \to Y \to Y' = X \to X' \to Y'$), thene there exists a (not necessarily unique) map $Z \to Z'$ which completes what we've got to a morphism of triangles.

Is there a criterion which ensures the uniqueness of this cone-map?

I'd like something along the lines of: if $\operatorname{Ext}^{-1}(X,Y')=0$ then yes.

(I might be too optimistic, cfr. Prop 10.1.17 of Kashiwara-Schapira Categories and Sheaves: in addition to $\operatorname{Hom}{(X[1],Y')} = 0$ they also assume $\operatorname{Hom} {(Y,X')} =0$. I really don't have this second assumption.)

(In the case I'm interested in $X=X', Y=Y'$ and $X\to X'$, $Y \to Y'$ are the identity maps.)

(If it makes things easier, although I doubt it, you can take the category to be the bounded derived category of coherent sheaves on some, fairly nasty, scheme.)

In the context I have in mind $X, Y, X', Y'$ are all objects of the heart of a bounded t-structure. If we assumed $\operatorname{Hom}{(Z,Y')} = 0$ or $\operatorname{Hom}{(X[1],Z')} = 0$ then the result easily follows. I don't think I'm happy making those assumptions though.

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    After all, you have the following weak functoriality in general: the mapping cone and the mapping cone triangle are unique up to non-canonical isomorphism.2011-02-25
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    I can't access KS at the moment. Is Prop 10.1.17 the BBD-criterion $\operatorname{Hom}(X[1],Z') =0$? I would also appreciate some hint why you think that $\operatorname{Hom}{(X[1],Y')}=0$ should help, I can't fit this information into any exact sequence relevant to your question.2011-02-25
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    I'll make some edits.2011-02-25
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    @Theo: In KS they use the Hom(Z,-) exact sequence to find a morphism $Z \to Y'$, the Hom(-,Z') exact sequence to find a morphism $X[1] \to Z'$. These morphisms satisfy $Z \to Y' \to Z' = Z \to X[1] \to Z'$. By axioms of triang'd cats we complete to morphism of triangles via map $Y[1] \to X'[1]$. This map has to be zero by KS's assumptions. By applying again Hom exact sequences we get a map $X[1]\to Y$, which has to be zero by assumption, thus the map we had $X[1]\to Z'$ is zero, thus $Z\to Z'$ is zero. (Yes, without drawing diagrams it's pretty incomprehensible.) Thanks for your interest guys.2011-02-25

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