Lemma 59.55.4. Consider a cartesian square
of schemes with $f$ an integral 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. The question is local on $Y$ and hence we may assume $Y$ is affine. Then we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with $f_ i : X_ i \to Y$ finite (this is easy in the affine case, but see Limits, Lemma 32.7.3 for a reference). Denote $p_{i'i} : X_{i'} \to X_ i$ the transition morphisms and $p_ i : X \to X_ i$ the projections. Setting $\mathcal{F}_ i = p_{i, *}\mathcal{F}$ we obtain from Lemma 59.51.9 a system $(\mathcal{F}_ i, \varphi _{i'i})$ with $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$. We get $f_*\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f_{i, *}\mathcal{F}_ i$ from Lemma 59.51.7. Set $X'_ i = Y' \times _ Y X_ i$ with projections $f'_ i$ and $g'_ i$. Then $X' = \mathop{\mathrm{lim}}\nolimits X'_ i$ as limits commute with limits. Denote $p'_ i : X' \to X'_ i$ the projections. We have
as desired. For the first equality see above. For the second use that pullback commutes with colimits. For the third use the finite case, see Lemma 59.55.3. For the fourth use Lemma 59.51.7. For the fifth use that $g'_ i \circ p'_ i = p_ i \circ g'$. For the sixth use that pullback commutes with colimits. For the seventh use $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$. $\square$
Your email address will not be published.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)
There are also: