The Stacks project

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

\begin{align*} \mathop{\mathrm{Hom}}\nolimits (Y, S(X)[1])^\vee & = \mathop{\mathrm{Hom}}\nolimits (Y[-1], S(X))^\vee \\ & = \mathop{\mathrm{Hom}}\nolimits (X, Y[-1]) \\ & = \mathop{\mathrm{Hom}}\nolimits (X[1], Y) \\ & = \mathop{\mathrm{Hom}}\nolimits (Y, S(X[1]))^\vee \end{align*}

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

\[ \ldots \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(B)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \to \mathop{\mathrm{Hom}}\nolimits (X, S(A)[1]) \to \ldots \]

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

\[ \ldots \leftarrow \mathop{\mathrm{Hom}}\nolimits (A, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (B, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (C, X) \leftarrow \mathop{\mathrm{Hom}}\nolimits (A[1], X) \leftarrow \ldots \]

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

\[ \xymatrix{ S(C)[-1] \ar[r]_-{S(h[-1])} & S(A) \ar[r]_{S(f)} & S(B) \ar[r]_{S(g)} & S(C) \ar[r]_{S(h)} & S(A)[1] \\ S(C)[-1] \ar[r]^-{S(h[-1])} \ar@{=}[u] & S(A) \ar[r]^ i \ar@{=}[u] & E \ar[r]^ p \ar@{..>}[u]^\varphi & S(C) \ar[r]^{S(h)} \ar@{=}[u] & S(A)[1] \ar@{=}[u] } \]

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

\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits (X, S(C)[-1]) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(B)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)[1]) \\ \mathop{\mathrm{Hom}}\nolimits (X, S(C)[-1]) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, E) \ar[r] \ar[u]^\varphi & \mathop{\mathrm{Hom}}\nolimits (X, S(C)) \ar[r] \ar@{=}[u] & \mathop{\mathrm{Hom}}\nolimits (X, S(A)[1]) \ar@{=}[u] } \]

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

\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits (E, S(C)) \ar[r] & \mathop{\mathrm{Hom}}\nolimits (S(A), S(C)) \\ \mathop{\mathrm{Hom}}\nolimits (E, S(B)) \ar[u] \ar[r] & \mathop{\mathrm{Hom}}\nolimits (S(A), S(B)) \ar[u] } \]

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

\[ \xymatrix{ \mathop{\mathrm{Hom}}\nolimits (C, E) \ar[d] & \mathop{\mathrm{Hom}}\nolimits (C, S(A)) \ar[l] \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits (B, E) & \mathop{\mathrm{Hom}}\nolimits (B, S(A)) \ar[l] } \]

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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