Say we have a cohomological functor F from a triangulated category $C$ to the category $Ab$ of abelian groups, e.g. $F=Hom(x,-)$, where x is an object in $C$. By definition, such a functor transform exact triangles into long exact sequence. And $F(y[i])$ is like the i-th "cohomology group" of the object $y$ w.r.t. the functor F. However, according to the philosophy of derived category, the right thing to look at is the complex, instead of the "cohomology groups". My question is, could there be a complex that "computes" these "cohomology groups"? e.g. Given two objects $x, y$ in $C$, can we find a complex that's like $RHom(x,y)$ whose cohomology computes $Ext_C^i(x,y):=Hom_C(x,y[i])$?

I would guess in full generality the answer is negative. (Otherwise it should have been done by Verdier.) But is there any mild or restrictive assumption that makes this true?

Or just for cohomological functors of the form $Hom_C(x,-)$, can we make the triangulated category $C$ enriched over $D^b(Ab)$ (such that the cohomology of the $Hom$ complex computes the old $Ext^i$'s?)


When I was writing this I feel like triangulated categories might not a good place to play game like this... If the above question isn't too interesting, could anyone just tell me what's the nature playground of questions like this? [Do dg-categories or $A_\infty$-category have an advantage on this?] Thanks.

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    $\begingroup$ Stable $(\infty,1)$-categories... $\endgroup$
    – David Roberts
    Apr 14, 2011 at 6:12

1 Answer 1


First, consider the category $\mathcal{S}$ of spectra. Let $S/2$ denote the cofibre of twice the identity map of the sphere spectrum. It is then known that the identity map of $S/2$ has order $4$. If we had a good system of chain complexes $Hom_{\mathcal{S}}(x,y)$ then $Hom_{\mathcal{S}}(S/2,S/2)$ would be the cofibre of twice the identity on $Hom_{\mathcal{S}}(S/2,S)$, but for chain complexes the cofibre of twice the identity always has exponent 2, which gives a contradiction. Thus, for triangulated categories of topological origin, the best you can hope for is to define mapping spectra $F(x,y)$, not chain complexes $Hom(x,y)$.

Next, this paper:

exhibits a triangulated category $\mathcal{C}$ such there are no nontrivial exact functors from $\mathcal{C}$ to any of the standard examples of triangulated categories, or in the opposite direction. This means in particular that there is no way to define a 'representable' functor $F_{\mathcal{C}}(x,-)$.

However, if $\mathcal{D}$ is a triangulated category arising from a stable model category then one can define mapping spectra $F_{\mathcal{D}}(x,y)$ for objects $x,y\in\mathcal{D}$. I think that is covered by this paper:

As in David Roberts's comment, I think one can also do the same for triangulated categories that arise from stable $(\infty,1)$-categories.

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    $\begingroup$ What does good mean here? Or, in other words, is there a notion of triangulated category enrichend over a triangulated category that includes a compatibility condition for cofibers? $\endgroup$
    – Rasmus
    Aug 1, 2012 at 8:16

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