Lemma 57.3.3. In the situation of Definition 57.3.2. If a Serre functor exists, then it is unique up to unique isomorphism and it is an exact functor of triangulated categories.
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
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