The Stacks project

[Proposition 2.16, Orlov-K3]; the fact that we do not need to assume vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ q(N, X)$ for $q > 0$ in the definition of negative objects above is due to [Canonaco-Stellari].

Proposition 57.11.6. Let $F$ and $F'$ be siblings as in Definition 57.11.1. Assume that $F$ is fully faithful and that $\mathcal{A}$ has enough negative objects (see above). Then $F$ and $F'$ are isomorphic functors.

Proof. By part (2) of Definition 57.11.1 the image of the functor $F'$ is contained in the essential image of the functor $F$. Hence the functor $H = F^{-1} \circ F'$ is a sibling of the identity functor. This reduces us to the case described in the next paragraph.

Let $\mathcal{D} = D^ b(\mathcal{A})$. We have to show a sibling $F : \mathcal{D} \to \mathcal{D}$ of the identity functor is isomorphic to the identity functor. Given an object $X$ of $\mathcal{D}$ let us say $X$ has width $w = w(X)$ if $w \geq 0$ is minimal such that there exists an integer $a \in \mathbf{Z}$ with $H^ i(X) = 0$ for $i \not\in [a, a + w - 1]$. Since $F$ is a sibling of the identity and since $F \circ [n] = [n] \circ F$ we are aready given isomorphisms

\[ c_ X : X \to F(X) \]

for $w(X) \leq 1$ compatible with shifts. Moreover, if $X = A[-a]$ and $X' = A'[-a]$ for some $A, A' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ then for any morphism $f : X \to X'$ the diagram

57.11.6.1
\begin{equation} \label{equiv-equation-to-show} \vcenter { \xymatrix{ X \ar[d]_{c_ X} \ar[r]_ f & X' \ar[d]^{c_{X'}} \\ F(X) \ar[r]^{F(f)} & F(X') } } \end{equation}

is commutative.

Next, let us show that for any morphism $f : X \to X'$ with $w(X), w(X') \leq 1$ the diagram (57.11.6.1) commutes. If $X$ or $X'$ is zero, this is clear. If not then we can write $X = A[-a]$ and $X' = A'[-a']$ for unique $A, A'$ in $\mathcal{A}$ and $a, a' \in \mathbf{Z}$. The case $a = a'$ was discussed above. If $a' > a$, then $f = 0$ (Derived Categories, Lemma 13.27.3) and the result is clear. If $a' < a$ then $f$ corresponds to an element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^ q(A, A')$ with $q = a - a'$. Using Yoneda extensions, see Derived Categories, Section 13.27, we can find $A = A_0, A_1, \ldots , A_{q - 1}, A_ q = A' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and elements

\[ \xi _ i \in \mathop{\mathrm{Ext}}\nolimits ^1(A_{i - 1}, A_ i) \]

such that $\xi $ is the composition $\xi _ q \circ \ldots \circ \xi _1$. In other words, setting $X_ i = A_ i[-a + i]$ we obtain morphisms

\[ X = X_0 \xrightarrow {f_1} X_1 \to \ldots \to X_{q - 1} \xrightarrow {f_ q} X_ q = X' \]

whose compostion is $f$. Since the commutativity of (57.11.6.1) for $f_1, \ldots , f_ q$ implies it for $f$, this reduces us to the case $q = 1$. In this case after shifting we may assume we have a distinguished triangle

\[ A' \to E \to A \xrightarrow {f} A'[1] \]

Observe that $E$ is an object of $\mathcal{A}$. Consider the following diagram

\[ \xymatrix{ E \ar[d]_{c_ E} \ar[r] & A \ar[d]_{c_ A} \ar[r]_ f & A'[1] \ar[d]^{c_{A'}[1]} \ar@{..>}@<-1ex>[d]_\gamma \ar@{..>}[ld]^\epsilon \ar[r] & E[1] \ar[d]^{c_ E[1]} \\ F(E) \ar[r] & F(A) \ar[r]^{F(f)} & F(A')[1] \ar[r] & F(E)[1] } \]

whose rows are distinguished triangles. The square on the right commutes already but we don't yet know that the middle square does. By the axioms of a triangulated category we can find a morphism $\gamma $ which does make the diagram commute. Then $\gamma - c_{A'}[1]$ composed with $F(A')[1] \to F(E)[1]$ is zero hence we can find $\epsilon : A'[1] \to F(A)$ such that $\gamma - c_{A'}[1] = F(f) \circ \epsilon $. However, any arrow $A'[1] \to F(A)$ is zero as it is a negative ext class between objects of $\mathcal{A}$. Hence $\gamma = c_{A'}[1]$ and we conclude the middle square commutes too which is what we wanted to show.

To finish the proof we are going to argue by induction on $w$ that there exist isomorphisms $c_ X : X \to F(X)$ for all $X$ with $w(X) \leq w$ compatible with all morphisms between such objects. The base case $w = 1$ was shown above. Assume we know the result for some $w \geq 1$.

Let $X$ be an object with $w(X) = w + 1$. Pick $a \in \mathbf{Z}$ with $H^ i(X) = 0$ for $i \not\in [a, a + w]$. Set $b = a + w$ so that $H^ b(X)$ is nonzero. Choose $N[-b] \to X$ as in Lemma 57.11.4. Choose a distinguished diagram

\[ N[-b] \to X \to Y \to N[-b + 1] \]

Computing the long exact cohomology sequence we find $w(Y) \leq w$. Hence by induction we find the solid arrows in the following diagram

\[ \xymatrix{ N[-b] \ar[r] \ar[d]_{c_ N[-b]} & X \ar[r] \ar@{..>}[d]_{c_{N[-b] \to X}} & Y \ar[r] \ar[d]^{c_ Y} & N[-b + 1] \ar[d]^{c_ N[-b + 1]} \\ F(N)[-b] \ar[r] & F(X) \ar[r] & F(Y) \ar[r] & F(N)[-b + 1] } \]

We obtain the dotted arrow $c_{N[-b] \to X}$. By Derived Categories, Lemma 13.4.8 the dotted arrow is unique because $\mathop{\mathrm{Hom}}\nolimits (X, F(N)[-b]) \cong \mathop{\mathrm{Hom}}\nolimits (X, N[-b]) = 0$ by our choice of $N$. In fact, $c_{N[-b] \to X}$ is the unique dotted arrow making the square with vertices $X, Y, F(X), F(Y)$ commute.

Let $N'[-b] \to X$ be another map as in Lemma 57.11.4 and let us prove that $c_{N[-b] \to X} = c_{N'[-b] \to X}$. Observe that the map $(N \oplus N')[-b] \to X$ also satisfies the conditions of Lemma 57.11.4. Thus we may assume $N'[-b] \to X$ factors as $N'[-b] \to N[-b] \to X$ for some morphism $N' \to N$. Choose distinguished triangles $N[-b] \to X \to Y \to N[-b + 1]$ and $N'[-b] \to X \to Y' \to N'[-b + 1]$. By axiom TR3 we can find a morphism $g : Y' \to Y$ which joint with $\text{id}_ X$ and $N' \to N$ forms a morphism of triangles. Since we have (57.11.6.1) for $g$ we conclude that

\[ (F(X) \to F(Y)) \circ c_{N'[-b] \to X} = (F(X) \to F(Y)) \circ c_{N[-b] \to X} \]

The uniqueness of $c_{N[-b] \to X}$ pointed out in the construction above now shows that $c_{N'[-b] \to X} = c_{N[-b] \to X}$.

Thus we can now define for $X$ of width $w + 1$ the isomorphism $c_ X : X \to F(X)$ as the common value of the maps $c_{N[-b] \to X}$ where $N[-b] \to X$ is as in Lemma 57.11.4. To finish the proof, we have to show that the diagrams (57.11.6.1) commute for all morphisms $f : X \to X'$ between objects with $w(X) \leq w + 1$ and $w(X') \leq w + 1$. Choose $a \leq b \leq a + w$ such that $H^ i(X) = 0$ for $i \not\in [a, b]$ and $a' \leq b' \leq a' + w$ such that $H^ i(X') = 0$ for $i \not\in [a', b']$. We will use induction on $(b' - a') + (b - a)$ to show the claim. (The base case is when this number is zero which is OK because $w \geq 1$.) We distinguish two cases.

Case I: $b' < b$. In this case, by Lemma 57.11.5 we may choose $N[-b] \to X$ as in Lemma 57.11.4 such that the composition $N[-b] \to X \to X'$ is zero. Choose a distuiguished triangle $N[-b] \to X \to Y \to N[-b + 1]$. Since $N[-b] \to X'$ is zero, we find that $f$ factors as $X \to Y \to X'$. Since $H^ i(Y)$ is nonzero only for $i \in [a, b - 1]$ we see by induction that (57.11.6.1) commutes for $Y \to X'$. The diagram (57.11.6.1) commutes for $X \to Y$ by construction if $w(X) = w + 1$ and by our first induction hypothesis if $w(X) \leq w$. Hence (57.11.6.1) commutes for $f$.

Case II: $b' \geq b$. In this case we choose $N'[-b'] \to X'$ as in Lemma 57.11.4. We may also assume that $\mathop{\mathrm{Hom}}\nolimits (H^{b'}(X), N') = 0$ (this is relevant only if $b' = b$), for example because we can replace $N'$ by an object $N''$ which surjects onto $N' \oplus H^{b'}(X)$ and such that $\mathop{\mathrm{Hom}}\nolimits (N' \oplus H^{b'}(X), N'') = 0$. We choose a distinguished triangle $N'[-b'] \to X' \to Y' \to N'[-b' + 1]$. Since $\mathop{\mathrm{Hom}}\nolimits (X, X') \to \mathop{\mathrm{Hom}}\nolimits (X, Y')$ is injective by our choice of $N'$ (details omitted) the same is true for $\mathop{\mathrm{Hom}}\nolimits (X, F(X')) \to \mathop{\mathrm{Hom}}\nolimits (X, F(Y'))$. Hence it suffices in this case to check that (57.11.6.1) commutes for the composition $X \to Y'$ of the morphisms $X \to X' \to Y'$. Since $H^ i(Y')$ is nonzero only for $i \in [a', b' - 1]$ we conclude by induction hypothesis. $\square$


Comments (2)

Comment #7172 by Noah Olander on

Typo: "Let be another" should say .


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