# Are functor categories with triangulated codomains themselves triangulated?

I'm fairly confident that the following assertion is true (but I will confess that I did not verify the octahedral axiom yet):

Let $$T$$ be a triangulated category and $$C$$ any category (let's say small to avoid alarming my set theorist friends). Then, the category of functors $$C \to T$$ inherits a natural triangulated structure from T.

By "natural" and "inherits" I mean that the shift map $$[1]$$ on our functor category sends each $$F:C \to T$$ to the functor $$F[1]$$ satisfying $$F[1](c) = F(c)[1]$$ on each object $$c$$ of $$C$$; and similarly, distinguished triangles of functors $$F \to G \to H \to F[1]$$ are precisely the ones for which over each object $$c$$ of $$C$$ we have a distinguished triangle in $$T$$ of the form $$F(c) \to G(c) \to H(c) \to F[1](c).$$

The main question is whether this has been written up in some standard book or paper (I couldn't find it in Gelfand-Manin for instance). Perhaps it is considered too obvious and relegated to an elementary exercise. Mostly, I am interested in inheriting t-structures and hearts from $$T$$ to functor categories $$C \to T$$, and would appreciate any available reference which deals with such matters.

• I don't think that this will work, essentially since the mapping cone does not define a functor from arrows in the homotopy category to the homotopy category. Compare also the discussion here: mathoverflow.net/questions/57904/… You might be interested in the notion of a stable derivator, which essentially assigns triangulated "functor categories" to all diagram shapes in a consistent way: arxiv.org/abs/1112.3840 – Bertram Arnold Dec 16 '20 at 22:49
• Thanks @BertramArnold; I had initially hoped that we wouldn't need to say something for all possible diagram shapes, only the C-indexed ones, but it seems that triangulated categories are much stranger than I'd initially thought. – Vidit Nanda Dec 16 '20 at 23:18

The statement is false.

For example, take $$C=[1]\times [1]$$ to be a square and $$\mathcal{T} = h\mathsf{Sp}$$ to be the homotopy category of spectra. Now consider the square $$X$$ with $$X(0,0) = S^2$$, $$X(1,0) = S^1$$, and the other values zero, and the other square $$Y$$ with $$Y(1,0) = S^1$$ and $$Y(1,1) = S^0$$. Take the maps $$S^2 \to S^1$$ and $$S^1 \to S^0$$ to be $$\eta$$, and consider the natural transformation $$X \to Y$$ which is given by multiplication by 2 on $$X(1,0)=S^1 \to S^1 = Y(1,0)$$.

If this map had a cofiber, then, from the initial to final vertex we would get a map $$S^3 \to S^0$$. Following the square one direction, we see that we would have some representative for the Toda bracket $$\langle \eta, 2, \eta\rangle$$. Following the other direction, we factor through zero. But this Toda bracket consists of the classes $$2\nu$$ and $$-2\nu$$; in particular, it does not contain zero.

[Of course, this example can be generalized to any nontrivial Toda bracket/Massey product in any triangulated category you're more familiar with.]

Indeed, the Toda bracket is exactly the obstruction to 'filling in the cube' for the natural transformation $$X \to Y$$.

Anyway- this is one of many reasons to drop triangulated categories in favor of one of the many modern alternatives (e.g. stable $$\infty$$-categories, derivators, etc.).

As for t-structures and so on, in the land of stable $$\infty$$-categories these are easy to come by. (See, e.g., Higher Algebra section 1.2.1 and Proposition 1.4.4.11 for various tricks for building these.)

Dylan Wilson's example is excellent. Let me offer another one, with a more algebraic and "finitistic" flavor.

In my opinion, the simplest triangulated category $$\mathcal{T}$$ is the category of finite-dimensional vector spaces over a field $$k$$, with identity suspension (a.k.a. translation) functor and $$3$$-periodic long exact sequences as exact triangles. (This is actually the only triangulated structure carried by $$\mathcal{T}$$ up to equivalence.)

Let $$C_2$$ be the cyclic group of order $$2$$ (regarded as a category with just one object). Then the functor category $$\mathcal{T}^{C_2}$$ is the category of finitely generated modules over the group algebra $$k[C_2]$$. This is the same as the category of finitely generated projective modules over the so-called Auslander algebra $$B$$ of $$k[C_2]$$. By a result Freyd, if $$\mathcal{T}^{C_2}$$ were triangulated then $$B$$ would be self-injective.

If $$k$$ has characteristic $$2$$, $$k[C_2]\cong k[\epsilon]/(\epsilon^2)$$ is the algebra of dual numbers and $$B$$ is the endomorphism algebra of the $$k[\epsilon]/(\epsilon^2)$$-module $$k\oplus k[\epsilon]/(\epsilon^2)$$. This $$B$$ is not self-injective. Indeed, since $$k$$ has characteristic $$2$$, $$k[\epsilon]/(\epsilon^2)$$ is not semi-simple, so $$B$$ has global dimension $$2$$. If $$B$$ were self-injective it would have global dimension either $$0$$ or $$\infty$$.

I believe I have a simpler counterexample, which I learned from Paul Balmer's course on tensor-triangular geometry last spring:

Claim The arrow category $$\mathcal{T}^{\bullet \to \bullet}$$ of a triangulated category $$\mathcal{T}$$ never has any triangulated structure unless $$\mathcal{T} = 0$$. Actually, we don't even need $$\mathcal{T}$$ to be triangulated here: if $$\mathcal{T}$$ is any additive category such that $$\mathcal{T}^{\bullet \to \bullet}$$ is triangulated, then $$\mathcal{T} = 0$$.

Proof: Suppose $$\mathcal{T}$$ is an additive category such that $$\mathcal{T}^{\bullet \to \bullet}$$ is triangulated. Let $$a$$ be an arbitrary object in $$\mathcal{T}$$, with identity morphism $$1_a : a \to a$$. Let $$t$$ denote the unique morphism $$a \to 0$$. Then $$\require{AMScd}$$ $$\begin{CD} a @>1_a>> a\\ @V 1_a V V @VV t V\\ a @>>t> 0 \end{CD}$$ defines a morphism $$\alpha : 1_a \to t$$ in $$\mathcal{T}^{\bullet \to \bullet}$$. Note that $$\alpha$$ is an epimorphism. In any triangulated category, all epimorphisms are split, so let $$\beta : t \to 1_a$$ be a splitting of $$\alpha$$ (that is, $$\alpha \circ \beta$$ is the identity morphism of $$t$$). Then $$\beta$$ is a commutative diagram $$\begin{CD} a @>t>> 0\\ @V f V V @VVs V\\ a @>>1_a> a \end{CD}$$ such that $$1_a \circ f = 1_a$$ (and $$t \circ s = 1_0$$). From this and the commutativity of the diagram, we see that $$1_a = 1_a \circ f = s \circ t$$ factors through $$0$$. Thus, $$a = 0$$. Since $$a$$ was arbitrary, $$\mathcal{T} = 0$$.

Edit: Of course we could make the statement even weaker: we only really needed that $$\mathcal{T}$$ has a zero object. But if $$\mathcal{T}^{\bullet \to \bullet}$$ is triangulated, then $$\mathcal{T}$$ must be additive, because it embeds as an additive subcategory of $$\mathcal{T}^{\bullet \to \bullet}$$ via $$a \mapsto 1_a$$.

• This example is amazing! – Fernando Muro Dec 19 '20 at 16:18