Lemma 56.13.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{D}$ be a triangulated category. Let $F, F' : D^ b(\mathcal{A}) \longrightarrow \mathcal{D}$ be exact functors of triangulated categories. Assume

the functors $F \circ i$ and $F' \circ i$ are isomorphic where $i : \mathcal{A} \to D^ b(\mathcal{A})$ is the inclusion functor, and

for all $X, Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we have $\mathop{\mathrm{Ext}}\nolimits ^ q_\mathcal {D}(F(X), F(Y)) = 0$ for $q < 0$ (for example if $F$ is fully faithful).

Then $F$ and $F'$ are siblings.

**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

\[ M_ i = (A^{-i} \to \ldots \to A^0)[-i],\quad i = 0, \ldots , n \]

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 13.40.2. Since both $F$ and $F'$ are exact functors of triangulated categories both

\[ F(M_ i) \quad \text{and}\quad F'(M_ i) \]

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

\[ F(A^{-n}) \to \ldots \to F(A^0) = F'(A^{-n}) \to \ldots \to F'(A^0) \]

Since all negative $\mathop{\mathrm{Ext}}\nolimits $s between these objects vanish by assumption we conclude by uniqueness of Postnikov systems (Derived Categories, Lemma 13.40.6) that $F(K) = F(M_ n[n]) \cong F'(M_ n[n]) = F'(K)$.
$\square$

## Comments (0)