The Stacks project

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

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

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

Proof. Given a Serre functor $S$ the object $S(X)$ represents the functor $Y \mapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee$. Thus the object $S(X)$ together with the functorial identification $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, Y)^\vee = \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(Y, S(X))$ is determined up to unique isomorphism by the Yoneda lemma (Categories, Lemma 4.3.5). Moreover, for $\varphi : X \to X'$, the arrow $S(\varphi ) : S(X) \to S(X')$ is uniquely determined by the corresponding transformation of functors $\mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X, -)^\vee \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {T}(X', -)^\vee$.

For objects $X, Y$ of $\mathcal{T}$ we have

By the Yoneda lemma we conclude that there is a unique isomorphism $S(X[1]) \to S(X)[1]$ inducing the isomorphism from top left to bottom right. Since each of the isomorphisms above is functorial in both $X$ and $Y$ we find that this defines an isomorphism of functors $S \circ [1] \to [1] \circ S$.

Let $(A, B, C, f, g, h)$ be a distinguished triangle in $\mathcal{T}$. We have to show that the triangle $(S(A), S(B), S(C), S(f), S(g), S(h))$ is distinguished. Here we use the canonical isomorphism $S(A[1]) \to S(A)[1]$ constructed above to identify the target $S(A[1])$ of $S(h)$ with $S(A)[1]$. We first observe that for any $X$ in $\mathcal{T}$ the triangle $(S(A), S(B), S(C), S(f), S(g), S(h))$ induces a long exact sequence

of finite dimensional $k$-vector spaces. Namely, this sequence is $k$-linear dual of the sequence

which is exact by Derived Categories, Lemma 13.4.2. Next, we choose a distinguished triangle $(S(A), E, S(C), i, p, S(h))$ which is possible by axioms TR1 and TR2. We want to construct the dotted arrow making following diagram commute

Namely, if we have $\varphi$, then we claim for any $X$ the resulting map $\mathop{\mathrm{Hom}}\nolimits (X, E) \to \mathop{\mathrm{Hom}}\nolimits (X, S(B))$ will be an isomorphism of $k$-vector spaces. Namely, we will obtain a commutative diagram

with exact rows (see above) and we can apply the 5 lemma (Homology, Lemma 12.5.20) to see that the middle arrow is an isomorphism. By the Yoneda lemma we conclude that $\varphi$ is an isomorphism. To find $\varphi$ consider the following diagram

The elements $p$ and $S(f)$ in positions $(0, 1)$ and $(1, 0)$ define a cohomology class $\xi$ in the total complex of this double complex. The existence of $\varphi$ is equivalent to whether $\xi$ is zero. If we take $k$-linear duals of this and we use the defining property of $S$ we obtain

Since both $A \to B \to C$ and $S(A) \to E \to S(C)$ are distinguished triangles, we know by TR3 that given elements $\alpha \in \mathop{\mathrm{Hom}}\nolimits (C, E)$ and $\beta \in \mathop{\mathrm{Hom}}\nolimits (B, S(A))$ mapping to the same element in $\mathop{\mathrm{Hom}}\nolimits (B, E)$, there exists an element in $\mathop{\mathrm{Hom}}\nolimits (C, S(A))$ mapping to both $\alpha$ and $\beta$. In other words, the cohomology of the total complex associated to this double complex is zero in degree $1$, i.e., the degree corresponding to $\mathop{\mathrm{Hom}}\nolimits (C, E) \oplus \mathop{\mathrm{Hom}}\nolimits (B, S(A))$. Taking duals the same must be true for the previous one which concludes the proof. $\square$