# 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 $$$$ on our functor category sends each $$F:C \to T$$ to the functor $$F$$ satisfying $$F(c) = F(c)$$ on each object $$c$$ of $$C$$; and similarly, distinguished triangles of functors $$F \to G \to H \to F$$ are precisely the ones for which over each object $$c$$ of $$C$$ we have a distinguished triangle in $$T$$ of the form $$F(c) \to G(c) \to H(c) \to F(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 Dec 16, 2020 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. Dec 16, 2020 at 23:18

The statement is false.

For example, take $$C=\times $$ 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.)

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! Dec 19, 2020 at 16:18

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 $$\infty$$.