8
votes
3answers
188 views

Is 'the' homotopy colimit of a sequence of exact triangles an exact triangle?

Let $$ \begin{array}{rccccl} A_0&\to& B_0&\to& C_0&\to\\ \downarrow & &\downarrow&&\downarrow\\ A_1&\to& B_1&\to& C_1&\to\\ \downarrow & ...
2
votes
2answers
192 views

Cool Examples of Localisation in Triangulated Cats Besides the Usual

In the theory of triangulated categories there is a hefty literature on localisation -- the most common example in algebra being (variants of) localising the homotopy category of chain complexes over ...
4
votes
1answer
203 views

Is the functor $mod(Tw\mathcal{C})\rightarrow mod(\mathcal{C})$ cohomologically full and faithful?

Let $\mathcal{C}$ be a $c$-unital $A_\infty$-category. If $\mathcal{A}$ is a $c$-unital and triangulated $A_\infty$-category, then there is a $c$-unital $A_\infty$- functor $$Tw: ...
5
votes
1answer
111 views

Is this a description of the $\aleph_1$-localizing subcategory generated by a compact generator?

This should be obvious but I'm not seeing it: The $\mathfrak T$ be a triangulated category with coproducts and with a compact generator $A$ (that is, the functor $\mathfrak T(A,\_)$ preserves ...
5
votes
2answers
349 views

Resolutions of unbounded complexes and homotopy (co)limits.

I want to understand once and for all what the resolution of an unbounded complex is. I've been trying to read 'Homotopy limits in triangulated categories' by Marcel Bokstedt and Amnon Neeman and ...
3
votes
0answers
299 views

Good morphisms of distinguished triangles: can Neeman's method be applied to the motivic stable homotopy category?

It is well known that non-uniqueness of a cone for a morphism in a triangulated category $C$ makes constructing exact functors (of triangulated categories) a difficult task. In section 3 of his "Some ...
8
votes
1answer
543 views

Cohomological functor from triangulated category

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 ...
15
votes
2answers
831 views

Why not define triangulated categories using a mapping cone functor?

Recall that the usual definition of a triangulated category is an additive category equipped with a self equivalence called $[1]$ in which certain diagrams, of the form $X \to Y \to Z \to X[1]$ are ...
9
votes
2answers
412 views

Does a triangulated category that possesses a subcategory $B$ of generators with no extensions of non-zero degree between them have to be isomorphic to $K^b(B)$?

Suppose that a triangulated category $C$ contains a full additive subcategory $B$ of (strong) generators (i.e. there does not exist a proper strict triangulated subcategory $C'\subset C$ that contains ...
1
vote
0answers
39 views

Any examples known of $K^b(B)$ localized by a set of morphisms (i.e. of complexes of length 1)?

I would like to understand the following setting: for an additive $B$ localize $K^b(B)$ by a set of $B$-morphisms (i.e. by a thick triangulated subcategory generated by some set of two-term ...
7
votes
4answers
688 views

Intuition about the triangulation of a homotopy category K(A)

Let $\cal{A}$ be an additive category. Given a morphism of (cochain) complexes $f:X\rightarrow Y$ we can form the mapping cone $C_f$, which is the complex $X[1]\oplus Y$ with differential given by ...
8
votes
1answer
640 views

How to write down the determinant of a quasi-isomorphism?

This question about the determinant of a perfect complex reminded me of an old question that I had. The construction of the determinant (as in MR1914072 or MR0437541) is a difficult piece of ...
13
votes
0answers
465 views

When is every “solid” perfect complex faithful?

Let $R$ be a noetherian commutative ring. Consider $D^{perf}(R)=K^b(R-proj)$ the category of bounded complexes of finitely generated projective $R$-modules, with maps of complexes up to homotopy. ...
11
votes
3answers
809 views

Classifying triangulated structures on a graded category

I know of several results to the effect that two triangulated categories are equivalent categories (usually one coming from algebra and one coming from topology). However, it's never been clear to me ...
14
votes
3answers
1k views

distinguished triangles and cohomology

Start with A an abelian category and form the derived category D(A). Take a triangle (not necessarily distinguished) and take it's cohomology. We obtain a long sequence (not necessarily exact). If the ...
6
votes
1answer
751 views

Sources for exact triangles in triangulated categories.

The other day I came across the statement that in the triangulated category $\mathfrak{KK}$ (of C*-algebras with KK-groups as morphism sets) "there are many other sources of exact triangles besides ...