Lemma 59.86.3. Let $I$ be a directed set. Consider an inverse system of cartesian diagrams of schemes
\[ \xymatrix{ X_ i \ar[d]_{f_ i} & Y_ i \ar[l]^{h_ i} \ar[d]^{e_ i} \\ S_ i & T_ i \ar[l]_{g_ i} } \]
with affine transition morphisms and with $g_ i$ quasi-compact and quasi-separated. Set $X = \mathop{\mathrm{lim}}\nolimits X_ i$, $S = \mathop{\mathrm{lim}}\nolimits S_ i$, $T = \mathop{\mathrm{lim}}\nolimits T_ i$ and $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ to obtain the cartesian diagram
\[ \xymatrix{ X \ar[d]_ f & Y \ar[l]^ h \ar[d]^ e \\ S & T \ar[l]_ g } \]
Let $(\mathcal{F}_ i, \varphi _{i'i})$ be a system of sheaves on $(T_ i)$ as in Definition 59.51.1. Set $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$ on $T$ where $p_ i : T \to T_ i$ is the projection. Then we have the following
If $f_ i^{-1}g_{i, *}\mathcal{F}_ i = h_{i, *}e_ i^{-1}\mathcal{F}_ i$ for all $i$, then $f^{-1}g_*\mathcal{F} = h_*e^{-1}\mathcal{F}$.
If $\mathcal{F}_ i$ is an abelian sheaf for all $i$ and $f_ i^{-1}R^ qg_{i, *}\mathcal{F}_ i = R^ qh_{i, *}e_ i^{-1}\mathcal{F}_ i$ for all $i$, then $f^{-1}R^ qg_*\mathcal{F} = R^ qh_*e^{-1}\mathcal{F}$.
Proof.
We prove (2) and we omit the proof of (1). We will use without further mention that pullback of sheaves commutes with colimits as it is a left adjoint. Observe that $h_ i$ is quasi-compact and quasi-separated as a base change of $g_ i$. Denoting $q_ i : Y \to Y_ i$ the projections, observe that $e^{-1}\mathcal{F} = \mathop{\mathrm{colim}}\nolimits e^{-1}p_ i^{-1}\mathcal{F}_ i = \mathop{\mathrm{colim}}\nolimits q_ i^{-1}e_ i^{-1}\mathcal{F}_ i$. By Lemma 59.51.8 this gives
\[ R^ qh_*e^{-1}\mathcal{F} = \mathop{\mathrm{colim}}\nolimits r_ i^{-1}R^ qh_{i, *}e_ i^{-1}\mathcal{F}_ i \]
where $r_ i : X \to X_ i$ is the projection. Similarly, we have
\[ f^{-1}Rg_*\mathcal{F} = f^{-1}\mathop{\mathrm{colim}}\nolimits s_ i^{-1}R^ qg_{i, *}\mathcal{F}_ i = \mathop{\mathrm{colim}}\nolimits r_ i^{-1}f_ i^{-1}R^ qg_{i, *}\mathcal{F}_ i \]
where $s_ i : S \to S_ i$ is the projection. The lemma follows.
$\square$
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