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

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 $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{T})$,

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

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