Lemma 57.10.5. 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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #7171 by Noah Olander on
Comment #7308 by Johan on