The Stacks project

Proof. Let $K \in D^ b(\mathcal{A})$. We will show $F(K)$ is isomorphic to $F'(K)$. We can represent $K$ by a bounded complex $A^\bullet $ of objects of $\mathcal{A}$. After replacing $K$ by a translation we may assume $A^ i = 0$ for $i > 0$. Choose $n \geq 0$ such that $A^{-i} = 0$ for $i > n$. The objects

form a Postnikov system in $D^ b(\mathcal{A})$ for the complex $A^\bullet = A^{-n} \to \ldots \to A^0$ in $D^ b(\mathcal{A})$. See Derived Categories, Example 13.41.2. Since both $F$ and $F'$ are exact functors of triangulated categories both

form a Postnikov system in $\mathcal{D}$ for the complex

Since all negative $\mathop{\mathrm{Ext}}\nolimits $s between these objects vanish by assumption we conclude by uniqueness of Postnikov systems (Derived Categories, Lemma 13.41.6) that $F(K) = F(M_ n[n]) \cong F'(M_ n[n]) = F'(K)$. $\square$

Proof. Part (1) is immediate from property (2) for siblings.

Assume $F$ is fully faithful. Denote $\mathcal{D}' \subset \mathcal{D}$ the essential image of $F$ so that $F : D^ b(\mathcal{A}) \to \mathcal{D}'$ is an equivalence. Since the functor $F'$ factors through $\mathcal{D}'$ by property (2) for siblings, we can consider the functor $H = F^{-1} \circ F' : D^ b(\mathcal{A}) \to D^ b(\mathcal{A})$. Observe that $H$ is a sibling of the identity functor. Since it suffices to prove that $H$ is fully faithful, we reduce to the problem discussed in the next paragraph.

Set $\mathcal{D} = D^ b(\mathcal{A})$. We have to show a sibling $F : \mathcal{D} \to \mathcal{D}$ of the identity functor is fully faithful. Denote $a_ X : X \to F(X)$ the functorial isomorphism for $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ given to us by Definition 57.10.1. For any $K$ in $\mathcal{D}$ and distinguished triangle $K_1 \to K_2 \to K_3$ of $\mathcal{D}$ if the maps

are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. This uses the $5$-lemma Homology, Lemma 12.5.20 and Derived Categories, Lemma 13.4.2; details omitted. Similarly, if the maps

are isomorphisms for all $n \in \mathbf{Z}$ and $i = 1, 3$, then the same is true for $i = 2$ and all $n \in \mathbf{Z}$. Using the canonical truncations and induction on the number of nonzero cohomology objects, we see that it is enough to show

is bijective for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and all $q \in \mathbf{Z}$. Since $F$ is a sibling of $\text{id}$ we have $F(X) \cong X$ and $F(Y) \cong Y$ hence the right hand side is zero for $q < 0$. The case $q = 0$ is OK by our assumption that $F$ is a sibling of the identity functor. It remains to prove the cases $q > 0$.

The case $q = 1$: Injectivity. An element $\xi $ of $\mathop{\mathrm{Ext}}\nolimits ^1(X, Y)$ gives rise to a distinguished triangle

Observe that $E \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. Since $F$ is a sibling of the identity functor we obtain a commutative diagram

whose vertical arrows are the isomorphisms $a_ E$ and $a_ X$. By TR3 the distinguished triangle associated to $\xi $ we started with is isomorphic to the distinguished triangle

Thus $\xi = 0$ if and only if $F(\xi )$ is zero, i.e., we see that $F : \mathop{\mathrm{Ext}}\nolimits ^1(X, Y) \to \mathop{\mathrm{Ext}}\nolimits ^1(F(X), F(Y))$ is injective.

The case $q = 1$: Surjectivity. Let $\theta $ be an element of $\mathop{\mathrm{Ext}}\nolimits ^1(F(X), F(Y))$. This defines an extension of $F(X)$ by $F(Y)$ in $\mathcal{A}$ which we may write as $F(E)$ as $F$ is a sibling of the identity functor. We thus get a distinguished triangle

for some morphisms $\alpha : Y \to E$ and $\beta : E \to X$. Since $F$ is a sibling of the identity functor, the sequence $0 \to Y \to E \to X \to 0$ is a short exact sequence in $\mathcal{A}$! Hence we obtain a distinguished triangle

for some morphism $\delta : X \to Y[1]$. Applying the exact functor $F$ we obtain the distinguished triangle

Arguing as above, we see that these triangles are isomorphic. Hence there exists a commutative diagram

for some isomorphisms $\gamma $, $\epsilon $ (we can say more but we won't need more information). We may write $\gamma = F(\gamma ')$ and $\epsilon = F(\epsilon ')$. Then we have $\theta = F(\epsilon ' \circ \delta \circ (\gamma ')^{-1})$ and we see the surjectivity holds.

The case $q > 1$: surjectivity. Using Yoneda extensions, see Derived Categories, Section 13.27, we find that for any element $\xi $ in $\mathop{\mathrm{Ext}}\nolimits ^ q(F(X), F(Y))$ we can find $F(X) = B_0, B_1, \ldots , B_{q - 1}, B_ q = F(Y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ and elements

such that $\xi $ is the composition $\xi _ q \circ \ldots \circ \xi _1$. Write $B_ i = F(A_ i)$ (of course we have $A_ i = B_ i$ but we don't need to use this) so that

by surjectivity for $q = 1$. Then $\eta = \eta _ q \circ \ldots \circ \eta _1$ is an element of $\mathop{\mathrm{Ext}}\nolimits ^ q(X, Y)$ with $F(\eta ) = \xi $.

The case $q > 1$: injectivity. An element $\xi $ of $\mathop{\mathrm{Ext}}\nolimits ^ q(X, Y)$ gives rise to a distinguished triangle

Applying $F$ we obtain a distinguished triangle

If $F(\xi ) = 0$, then $F(E) \cong F(Y)[q - 1] \oplus F(X)$ in $\mathcal{D}$, see Derived Categories, Lemma 13.4.11. Since $F$ is a sibling of the identity functor we have $E \cong F(E)$ and hence

In other words, $E$ is isomorphic to the direct sum of its cohomology objects. This implies that the initial distinguished triangle is split, i.e., $\xi = 0$. $\square$

Proof. Using the truncation functors we can represent $X$ by a complex $A^ a \to A^{a + 1} \to \ldots \to A^ b$ of objects of $\mathcal{A}$. Choose $N$ in $\mathcal{A}$ such that there exists a surjection $t : N \to A^ b$ and such that $\mathop{\mathrm{Hom}}\nolimits (A^ b, N) = 0$. Then the surjection $t$ defines a map $N[-b] \to X$ as desired. $\square$

Proof. We can represent $f$ by a map $f^\bullet : A^\bullet \to B^\bullet $ of bounded complexes of objects of $\mathcal{A}$, see for example Derived Categories, Lemma 13.11.6. Consider the object

of $\mathcal{A}$. Since $H^ b(B^\bullet ) = 0$ we see that $C \to H^ b(A^\bullet )$ is surjective. On the other hand, the map $C \to A^ b \to B^ b$ is the same as the map $C \to B^{b - 1} \to B^ b$ and hence the composition $C[-b] \to X \to X'$ is zero. Since $\mathcal{A}$ has enough negative objects, we can find an object $N$ which has a surjection $N \to C \oplus H^ b(X)$ such that $\mathop{\mathrm{Hom}}\nolimits (C \oplus H^ b(X), N) = 0$. Then $N$ together with the map $N[-b] \to X$ is a solution to the problem posed by the lemma. $\square$

Proof. By part (2) of Definition 57.10.1 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

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

is commutative.

Next, let us show that for any morphism $f : X \to X'$ with $w(X), w(X') \leq 1$ the diagram (57.10.6.1) 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

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

whose compostion is $f$. Since the commutativity of (57.10.6.1) 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

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

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 57.10.4. Choose a distinguished diagram

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

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 57.10.4 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 57.10.4. 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 (57.10.6.1) for $g$ we conclude that

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 57.10.4. To finish the proof, we have to show that the diagrams (57.10.6.1) 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 57.10.5 we may choose $N[-b] \to X$ as in Lemma 57.10.4 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 (57.10.6.1) commutes for $Y \to X'$. The diagram (57.10.6.1) commutes for $X \to Y$ by construction if $w(X) = w + 1$ and by our first induction hypothesis if $w(X) \leq w$. Hence (57.10.6.1) commutes for $f$.

Case II: $b' \geq b$. In this case we choose $N'[-b'] \to X'$ as in Lemma 57.10.4. 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 (57.10.6.1) 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. $\square$