The Stacks project

Lemma 57.87.8. Let $f : X \to Y$ and $g : Y \to Z$ be proper morphisms of schemes. Assume

  1. cohomology commutes with base change for $f$, and

  2. cohomology commutes with base change for $g$.

Then cohomology commutes with base change for $g \circ f$.

Proof. We will use the equivalence of Lemma 57.87.6 without further mention. Let $\ell $ be a prime number. Let $\mathcal{I}$ be an injective sheaf of $\mathbf{Z}/\ell \mathbf{Z}$-modules on $X_{\acute{e}tale}$. Then $f_*\mathcal{I}$ is an injective sheaf of $\mathbf{Z}/\ell \mathbf{Z}$-modules on $Y_{\acute{e}tale}$ (Cohomology on Sites, Lemma 21.14.2). The result follows formally from this, but we will also spell it out.

Let $Z' \to Z$ be a morphism of schemes and set $Y' = Z' \times _ Z Y$ and $X' = Z' \times _ Z X = Y' \times _ Y X$. Denote $a : X' \to X$, $b : Y' \to Y$, and $c : Z' \to Z$ the projections. Similarly for $f' : X' \to Y'$ and $g' : Y' \to Z'$. By Lemma 57.87.5 we have $b^{-1}f_*\mathcal{I} = f'_*a^{-1}\mathcal{I}$. On the other hand, we know that $R^ qf'_*a^{-1}\mathcal{I}$ and $R^ q(g')_*b^{-1}f_*\mathcal{I}$ are zero for $q > 0$. Using the spectral sequence (Cohomology on Sites, Lemma 21.14.7)

\[ R^ pg'_* R^ qf'_* a^{-1}\mathcal{I} \Rightarrow R^{p + q}(g' \circ f')_* a^{-1}\mathcal{I} \]

we conclude that $R^ p(g' \circ f')_*a^{-1}\mathcal{I} = 0$ for $p > 0$ as desired. $\square$


Comments (2)

Comment #3213 by Alexander Schmidt on

In the statement of the Lemma, should be replaced by .

There are also:

  • 4 comment(s) on Section 57.87: The proper base change theorem

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Your email address will not be published.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 0A4D. Beware of the difference between the letter 'O' and the digit '0'.