Lemma 57.3.1. Let $k$ be a field. Let $\mathcal{T}$ be a $k$-linear triangulated category such that $\dim _ k \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) < \infty$ for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$. The following are equivalent

1. there exists a $k$-linear equivalence $S : \mathcal{T} \to \mathcal{T}$ and $k$-linear isomorphisms $c_{X, Y} : \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))^\vee$ functorial in $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$,

2. for every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$ the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$ is representable and the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$ is corepresentable.

Proof. Condition (1) implies (2) since given $(S, c)$ and $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$ the object $S(X)$ represents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$ and the object $S^{-1}(X)$ corepresents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$.

Assume (2). We will repeatedly use the Yoneda lemma, see Categories, Lemma 4.3.5. For every $X$ denote $S(X)$ the object representing the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$. Given $\varphi : X \to X'$, we obtain a unique arrow $S(\varphi ) : S(X) \to S(X')$ determined by the corresponding transformation of functors $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, -)^\vee \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X', -)^\vee$. Thus $S$ is a functor and we obtain the isomorphisms $c_{X, Y}$ by construction. It remains to show that $S$ is an equivalence. For every $X$ denote $S'(X)$ the object corepresenting the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X)^\vee$. Arguing as above we find that $S'$ is a functor. We claim that $S'$ is quasi-inverse to $S$. To see this observe that

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(S'(S(X)), Y)$

bifunctorially, i.e., we find $S' \circ S \cong \text{id}_\mathcal {T}$. Similarly, we have

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, X) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(S'(X), Y)^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(S'(X)))$

and we find $S \circ S' \cong \text{id}_\mathcal {T}$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).