and
nonabelian homological algebra
Just as a derivator is a notion which lies in between a model category or a (∞,1)-category and its homotopy category, a stable (or triangulated) derivator is a notion which lies in between a stable model category or a stable (∞,1)-category and its homotopy triangulated category.
Stable derivators are a useful refinement of triangulated categories, since they contain enough information so that homotopy limit and colimit constructions can be performed “coherently” and desired maps and objects can be specified by true universal properties. This resolves many common issues with triangulated categories stemming from the fact that at the level of a homotopy category, certain desirable maps can only be stipulated to exist with some weak properties, but not characterized precisely among the class of maps that have those properties.
As is the case for stable (∞,1)-categories, being stable is a property of a derivator rather than a structure (or more precisely, it is a property-like structure). In fact, an $(\infty,1)$-category is stable precisely when its underlying derivator is.
Write $\square$ for the “free-living commutative square”
and let $L$ and $R$ denote the upper-left span and the lower-right cospan as subcategories of $\square$, with inclusions $i\colon L\to \square$ and $j \colon R\to \square$. For a derivator $D$, an object $X\in D(\square)$ is said to be cartesian if the unit $X\to j_* j^* X$ is an isomorphism, and dually cocartesian if the counit $i_! i^* X\to X$ is an isomorphism. Since $i$ and $j$ are fully faithful, it follows from a general theorem about derivators that $i_!$ and $j_*$ are also fully faithful; thus being cartesian is equivalent to being of the form $j_* Y$ for some $Y\in D(R)$, and likewise being cocartesian is equivalent to being of the form $i_! Z$.
A derivator $D$ is stable (or triangulated) if it is pointed, and moreover an $X\in D(\square)$ is cartesian if and only if it is cocartesian. Such a square is then called bicartesian.
One of the central facts about stable derivators is:
If $D$ is a stable derivator, then each category $D(X)$ is a triangulated category in a canonical way.
We describe the constructions when $X=1$. The shift/suspension functor $S\colon D(1)\to D(1)$ is defined by
where $a\colon 1\to L$ is the inclusion of the top-left vertex of the span, and $b\colon 1\to \square$ is the inclusion of the bottom-right vertex of the square. In other words, for an object $X\in D(1)$ its suspension is defined by the homotopy pushout
This makes sense in the more general context of any pointed derivator, but the stability axiom guarantees that $S$ is actually an equivalence of categories. Its inverse being given by the obvious dual “loop space” construction (which in a general pointed derivator is only right adjoint to it). This provides a motivation for the stability axiom: it is a generalization of the statement that every object is the loop space of its suspension and the suspension of its loop space.
One can also prove that $S$ is also a copower with the pointed circle $S^1$ in a suitable sense. In particular, since every object is isomorphic to a double suspension, it is a cogroup object; thus $D(X)$ is canonically an Ab-enriched category.
Let $Q$ denote the category
with inclusions $m,n,p\colon \square \to Q$ of the left and right squares and the outer rectangle, respectively. An object $X\in D(Q)$ is bicartesian if $m^*X$, $n^*X$, and $p^*X$ are all bicartesian.
Now consider a bicartesian object $X\in D(Q)$ of the form:
In other words, we stipulate that the restrictions to $D(1)$ along the inclusions of the lower-left and upper-right vertices be zero objects. Now since $X$ is bicartesian, the outer square
is bicartesian, and thus induces an isomorphism $D \cong S A$. Thus, from $X$ we can extract a “triangle”
and we define the distinguished triangles in $D(1)$ to be those isomorphic to triangles obtained in this way.
One can then prove the axioms of a triangulated category.
If $A \overset{f}{\to} B \overset{g}{\to} C \overset{h}{\to} S A$ is a distinguished triangle in a triangulated category, then $A \overset{f}{\to} B \overset{g}{\to} C \overset{-h}{\to} S A$ is also an “exact” triangle in the sense that it induces long exact sequences in homology and cohomology, but it is not in general distinguished. On the other hand, if
is bicartesian in an $(\infty,1)$-category, then so is
This seeming paradox is resolved by noticing that although these two bicartesian diagrams have the same object $D$ at their lower-right-hand corner, the different maps $h$ and $-h$ cause these diagrams to induce different isomorphisms $D \cong S A$. The isomorphism for the latter diagram incorporates an extra minus sign, relative to the first one, causing these two diagrams to both induce the same triangle $A \overset{f}{\to} B \overset{g}{\to} C \overset{h}{\to} S A$ in the homotopy category.
stable derivator
See derivator for general references about derivators, and also pointed derivator. References particularly pertaining to the stable version include:
Alex Heller, “Stable homotopy theories and stabilization” MR
Denis-Charles Cisinski and Amnon Neeman, “Additivity for derivator K-theory”, MR
Jens Franke, Uniqueness theorems for certain triangulated categories with an Adams spectral sequence, K-theory archive
Moritz Groth, Derivators, pointed derivators, and stable derivators (pdf)