TITLE: Are functor categories with triangulated codomains themselves triangulated?
QUESTION [20 upvotes]: 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.

REPLY [13 votes]: 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 $a \mapsto 1_a$.