[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 .

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

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 $N'[b] \to X$ be another" should say $N'[-b]$.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FZW. Beware of the difference between the letter 'O' and the digit '0'.