Lemma 57.11.5. Let $\mathcal{A}$ be an abelian category with enough negative objects. Let $f : X \to X'$ be a morphism of $D^ b(\mathcal{A})$. Let $b \in \mathbf{Z}$ such that $H^ i(X) = 0$ for $i > b$ and $H^ i(X') = 0$ for $i \geq b$. Then there exists a map $N[-b] \to X$ such that the induced map $N \to H^ b(X)$ is surjective, such that $\mathop{\mathrm{Hom}}\nolimits (H^ b(X), N) = 0$, and such that the composition $N[-b] \to X \to X'$ is zero.

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

$C = \mathop{\mathrm{Ker}}(A^ b \to A^{b + 1}) \times _{\mathop{\mathrm{Ker}}(B^ b \to B^{b + 1})} B^{b - 1}$

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. $\square$

Comment #7171 by Noah Olander on

The integer $a$ in the statement plays no role in the result or proof.

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