Lemma 94.20.3. Let $S$ be a scheme. Let $\tau \in \{ Zariski,\linebreak[0] {\acute{e}tale},\linebreak[0] smooth,\linebreak[0] syntomic,\linebreak[0] fppf\} $. Let

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

be a $2$-cartesian diagram of algebraic stacks over $S$. Then the base change map is an isomorphism

\[ g^{-1}Rf_*\mathcal{F} \longrightarrow Rf'_*(g')^{-1}\mathcal{F} \]

functorial for $\mathcal{F}$ in $\textit{Ab}(\mathcal{X}_\tau )$ or $\mathcal{F}$ in $\textit{Mod}(\mathcal{X}_\tau , \mathcal{O}_\mathcal {X})$.

**Proof.**
The isomorphism $g^{-1}f_*\mathcal{F} = f'_*(g')^{-1}\mathcal{F}$ is Lemma 94.5.1 (and it holds for arbitrary presheaves). For the derived direct images, there is a base change map because the morphisms $g$ and $g'$ are flat, see Cohomology on Sites, Section 21.15. To see that this map is a quasi-isomorphism we can use that for an object $y'$ of $\mathcal{Y}'$ over a scheme $V$ there is an equivalence

\[ (\mathit{Sch}/V)_{fppf} \times _{g(y'), \mathcal{Y}} \mathcal{X} = (\mathit{Sch}/V)_{fppf} \times _{y', \mathcal{Y}'} (\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}) \]

We conclude that the induced map $g^{-1}R^ if_*\mathcal{F} \to R^ if'_*(g')^{-1}\mathcal{F}$ is an isomorphism by Lemma 94.20.2.
$\square$

## Comments (2)

Comment #5830 by Georg Oberdieck on

Comment #5831 by Johan on