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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. –  Rasmus Feb 25 '11 at 13:59
    
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. –  t.b. Feb 25 '11 at 14:38
    
I'll make some edits. –  slackenerny Feb 25 '11 at 15:19
    
@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. –  slackenerny Feb 25 '11 at 15:30
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The uniqueness condition of the maps between cones is very restrictive. If it holds for every commutative square, this indeed means that you could define a "cone functor" $\mathrm{Mor}(\mathcal T) \to \mathcal T$ from the category of morphisms of your triangulated category $\mathcal T$ to $\mathcal T$ itself (just choose a cone object for each morphism, then your uniqueness condition ensures that the cone functor is well defined on morphisms). It turns out that this makes $\mathcal T$ a semisimple abelian category, if $\mathcal T$ is assumed to be Karoubian (i.e. every idempotent splits; many common triangulated categories are Karoubian). I found a proof of this claim in the following article:

http://www.math.uni-bielefeld.de/~gstevens/no_functorial_cones.pdf

In conclusion: you can't expect your uniqueness condition to hold globally in "useful" triangulated categories. There is a technique to overcome this problem, that is, using pre-triangulated dg-categories (introduced by Bondal and Kapranov) to "lift" triangulated categories. In this new framework you indeed have functorial cones.

Perhaps this doesn't answer your specific question (which, as I understand, is about a given commutative square), but it should point out that the desired uniqueness is, roughly speaking, very difficult to obtain in general.

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