Let $C$ be a triangulated category that is closed with respect to arbitrary small coproducts; let $D$ be some class of objects of $C$. Then it would be natural to say that $D$ generates $C$ either if

(i) There are no proper triangulated subcategories of $C$ that are closed with respect to small coproducts and contain $D$.

or if

(ii) For any non-zero object $c$ of $C$ there exist an integer $i$, a $d\in D$, and a non-zero $C$-morphism from $d[i]$ to $c$.

My question is: are there any 'well-known' relations between conditions (i) and (ii)? I only know a proof of their equivalence in the case when $D$ is a set of compact objects.


The obvious implication is (i) $\Longrightarrow$ (ii), of course. Indeed, the class of all $c\in C$ not satisfying (ii) is precisely the full triangulated subcategory of objects right orthogonal to the minimal triangulated subcategory satisfying (i) in $C$. Any triangulated subcategory with a nontrivial orthogonal subcategory is proper, hence the negation of (ii) implies the negation of (i).

In addition to the compactly generated case, an implication like (ii) implies (i) may also hold in the so-called well-generated case (as defined by Neeman in his book on triangulated categories and studied further by Krause, whose writings on this subject may be more accessible than Neeman's, as well as by later authors). The key word is "Brown representability" or more precisely the "contravariant Brown representability" in triangulated categories.

  • $\begingroup$ Thank you! Yes; I have met the equivalence of (i) and (ii) in the paper of Krause on well-generated triangulated categories. Yet I wonder whether well-generatedness is necessary here. $\endgroup$ Jul 25 '14 at 13:25
  • $\begingroup$ The argument goes like this: suppose that the Brown representability holds for the minimal triangulated subcategory $T$ in $C$ satisfying (i). Given an object $x\in C$, consider the functor $Hom_C({-},x)$ on the category $T$ and denote by $t\in T$ an object representing it. Then the identity morphism $t\to t$ corresponds to a certain morphism $t\to x$. The cone $c$ of this morphism satisfies the negation of (ii), so it must be zero. $\endgroup$ Jul 25 '14 at 13:53
  • $\begingroup$ Thank you! It seems that the implication you indicated is exactly what I need for my purposes. $\endgroup$ Jul 26 '14 at 19:56

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