Lemma 59.18.7. Notation and assumptions as in Definition 59.18.1. If $\mathcal{I}$ is an injective object of $\textit{PAb}(\mathcal{C})$, then $\check H^ p(\mathcal{U}, \mathcal{I}) = 0$ for all $p > 0$.
Proof. The Čech complex is the result of applying the functor $\mathop{\mathrm{Hom}}\nolimits _{\textit{PAb}(\mathcal{C})}(-, \mathcal{I}) $ to the complex $ \mathbf{Z}^\bullet _\mathcal {U} $, i.e.,
\[ \check H^ p(\mathcal{U}, \mathcal{I}) = H^ p (\mathop{\mathrm{Hom}}\nolimits _{\textit{PAb}(\mathcal{C})} (\mathbf{Z}^\bullet _\mathcal {U}, \mathcal{I})). \]
But we have just seen that $\mathbf{Z}^\bullet _\mathcal {U}$ is exact in negative degrees, and the functor $\mathop{\mathrm{Hom}}\nolimits _{\textit{PAb}(\mathcal{C})}(-, \mathcal{I})$ is exact, hence $\mathop{\mathrm{Hom}}\nolimits _{\textit{PAb}(\mathcal{C})} (\mathbf{Z}^\bullet _\mathcal {U}, \mathcal{I})$ is exact in positive degrees. $\square$
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Your email address will not be published.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 (0)