# exactness in triangulated categories is reflected by hom-functor

let $T$ be a triangulated category and $A \to B \to C \to A[1]$ a triangle in $T$ such that for every $A_0 \in T$ the induced long sequence

$... \to \hom(A_0,A) \to \hom(A_0,B) \to \hom(A_0,C) \to \hom(A_0,A[1]) \to \hom(A_0,B[1]) \to ...$

is exact. is then $A \to B \to C \to A[1]$ exact? I'm a beginner, so this could be rather trivial. I've checked it in special cases, but I can't translate the proof into $T$. I would like the result because it would imply that the homological algebra you know for abelian groups takes over to triangulated categories. comparable to the fact that you can work with group objects in arbitrary categories as with ordinary groups. you can do it directly with the axioms, but this is a big mess.

in many answers I've seen here so far, even standard facts are enriched with wonderful, ample insights. so you are also invited to dish durt about these basics of triangulated categories because I'm just beginning to learn them.

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no: consider changing the sign of one of your maps.

about your philosophical question concerning homological algebra for abelian groups carrying over to triangulated categories: actually the fundamental triangulated category is not of abelian groups, but of spectra (in algebraic topology). to make correct and precise statements illustrating this claim one should work with enhanced versions of triangulated categories, my favorite being stable (\infty,1)-categories in the sense of Lurie's DAG I. then for instance one can (probably?) say that any stable (\infty,1)-category is canonically enriched and tensored in spectra, formulate an analog of the mitchell embedding theorem, etc...

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Of course to be tensored over all spectra, you need a sufficient supply of colimits. But yes, this is very much the right point of view. –  Reid Barton Dec 30 '09 at 23:33