The Stacks project

Lemma 30.7.2. With notation as in diagram ( Assume $f : X \to S$ and $\mathcal{F}$ satisfy the hypotheses of Lemma 30.7.1. Choose a finite affine open covering $\mathcal{U} : X = \bigcup U_ i$ of $X$. There is a canonical isomorphism

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) \longrightarrow Rf'_*\mathcal{F}' \]

in $D^{+}(S')$. Moreover, if $S' \to S$ is affine, then in fact

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \]

with $\mathcal{U}' : X' = \bigcup U_ i'$ where $U_ i' = (g')^{-1}(U_ i) = U_{i, S'}$ is also affine.

Proof. In fact we may define $U_ i' = (g')^{-1}(U_ i) = U_{i, S'}$ no matter whether $S'$ is affine over $S$ or not. Let $\mathcal{U}' : X' = \bigcup U_ i'$ be the induced covering of $X'$. In this case we claim that

\[ g^*\check{\mathcal{C}}^\bullet (\mathcal{U}, f, \mathcal{F}) = \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \]

with $\check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}')$ defined in exactly the same manner as in Lemma 30.7.1. This is clear from the case of affine morphisms (Lemma 30.5.1) by working locally on $S'$. Moreover, exactly as in the proof of Lemma 30.7.1 one sees that there is an isomorphism

\[ \check{\mathcal{C}}^\bullet (\mathcal{U}', f', \mathcal{F}') \longrightarrow Rf'_*\mathcal{F}' \]

in $D^{+}(S')$ since the morphisms $U_ i' \to X'$ and $U_ i' \to S'$ are still affine (being base changes of affine morphisms). Details omitted. $\square$

Comments (0)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01XM. Beware of the difference between the letter 'O' and the digit '0'.