The Stacks project

Lemma 59.55.3. Consider a cartesian square

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y } \]

of schemes with $f$ a finite morphism. For any sheaf of sets $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $f'_*(g')^{-1}\mathcal{F} = g^{-1}f_*\mathcal{F}$.

Proof. In great generality there is a pullback map $g^{-1}f_*\mathcal{F} \to f'_*(g')^{-1}\mathcal{F}$, see Sites, Section 7.45. It suffices to check on stalks (Theorem 59.29.10). Let $\overline{y}' : \mathop{\mathrm{Spec}}(k) \to Y'$ be a geometric point. We have

\begin{align*} (f'_*(g')^{-1}\mathcal{F})_{\overline{y}'} & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} ((g')^{-1}\mathcal{F})_{\overline{x}'} \\ & = \prod \nolimits _{\overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'} \mathcal{F}_{g' \circ \overline{x}'} \\ & = \prod \nolimits _{\overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'} \mathcal{F}_{\overline{x}} \\ & = (f_*\mathcal{F})_{g \circ \overline{y}'} \\ & = (g^{-1}f_*\mathcal{F})_{\overline{y}'} \end{align*}

The first equality by Proposition 59.55.2. The second equality by Lemma 59.36.2. The third equality holds because the diagram is a cartesian square and hence the map

\[ \{ \overline{x}' : \mathop{\mathrm{Spec}}(k) \to X',\ f' \circ \overline{x}' = \overline{y}'\} \longrightarrow \{ \overline{x} : \mathop{\mathrm{Spec}}(k) \to X,\ f \circ \overline{x} = g \circ \overline{y}'\} \]

sending $\overline{x}'$ to $g' \circ \overline{x}'$ is a bijection. The fourth equality by Proposition 59.55.2. The fifth equality by Lemma 59.36.2. $\square$

Comments (0)

There are also:

  • 1 comment(s) on Section 59.55: Vanishing of finite higher direct images

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Your email address will not be published.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 0959. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 59.55.3 as pdf