**Proof.**
How is the map of the lemma constructed? For $i' \geq i$ we have a commutative diagram

\[ \xymatrix{ X \ar[r]_{f_{i'}} \ar[d]_ g & X_{i'} \ar[r]_{f_{i'i}} \ar[d]_{g_{i'}} & X_ i \ar[d]^{g_ i} \\ S \ar[r]^{h_{i'}} & S_{i'} \ar[r]^{h_{i'i}} & S_ i } \]

If we combine the base change map $h_{i'i}^{-1}Rg_{i, *}\mathcal{F}_ i \to Rg_{i', *}f_{i'i}^{-1}\mathcal{F}_ i$ (Cohomology on Sites, Lemma 21.15.1 or Remark 21.19.3) with the map $Rg_{i', *}\varphi _{i'i}$, then we obtain $\psi _{i'i} : h_{i' i}^{-1} R^ p g_{i, *} \mathcal{F}_ i \to R^ pg_{i', *} \mathcal{F}_{i'}$. Similarly, using the left square in the diagram we obtain maps $\psi _ i : h_ i^{-1}R^ pg_{i, *}\mathcal{F}_ i \to R^ pg_*\mathcal{F}$. The maps $h_{i'}^{-1}\psi _{i'i}$ and $\psi _ i$ are the maps used in the statement of the lemma. For this to make sense, we have to check that $\psi _{i''i} = \psi _{i''i'} \circ h_{i''i'}^{-1}\psi _{i'i}$ and $\psi _{i'} \circ h_{i'}^{-1}\psi _{i'i} = \psi _ i$; this follows from Cohomology on Sites, Remark 21.19.5.

Proof of the equality. First proof using dimension shifting^{1}. For any $U$ affine and étale over $X$ by Theorem 59.51.3 we have

\[ g_*\mathcal{F}(U) = H^0(U \times _ S X, \mathcal{F}) = \mathop{\mathrm{colim}}\nolimits H^0(U_ i \times _{S_ i} X_ i, \mathcal{F}_ i) = \mathop{\mathrm{colim}}\nolimits g_{i, *}\mathcal{F}_ i(U_ i) \]

where the colimit is over $i$ large enough such that there exists an $i$ and $U_ i$ affine étale over $S_ i$ whose base change is $U$ over $S$ (see Lemma 59.51.2). The right hand side is equal to $(\mathop{\mathrm{colim}}\nolimits h_ i^{-1}g_{i, *}\mathcal{F}_ i)(U)$ by Sites, Lemma 7.18.4. This proves the lemma for $p = 0$. If $(\mathcal{G}_ i, \varphi _{i'i})$ is a system with $\mathcal{G} = \mathop{\mathrm{colim}}\nolimits f_ i^{-1}\mathcal{G}_ i$ such that $\mathcal{G}_ i$ is an injective abelian sheaf on $X_ i$ for all $i$, then for any $U$ affine and étale over $X$ by Theorem 59.51.3 we have

\[ H^ p(U \times _ S X, \mathcal{G}) = \mathop{\mathrm{colim}}\nolimits H^ p(U_ i \times _{S_ i} X_ i, \mathcal{G}_ i) = 0 \]

for $p > 0$ (same colimit as before). Hence $R^ pg_*\mathcal{G} = 0$ and we get the result for $p > 0$ for such a system. In general we may choose a short exact sequence of systems

\[ 0 \to (\mathcal{F}_ i, \varphi _{i'i}) \to (\mathcal{G}_ i, \varphi _{i'i}) \to (\mathcal{Q}_ i, \varphi _{i'i}) \to 0 \]

where $(\mathcal{G}_ i, \varphi _{i'i})$ is as above, see Cohomology on Sites, Lemma 21.16.4. By induction the lemma holds for $p - 1$ and by the above we have vanishing for $p$ and $(\mathcal{G}_ i, \varphi _{i'i})$. Hence the result for $p$ and $(\mathcal{F}_ i, \varphi _{i'i})$ by the long exact sequence of cohomology.

Second proof. Recall that $S_{affine, {\acute{e}tale}} = \mathop{\mathrm{colim}}\nolimits (S_ i)_{affine, {\acute{e}tale}}$, see Lemma 59.51.2. Thus if $U$ is an object of $S_{affine, {\acute{e}tale}}$, then we can write $U = U_ i \times _{S_ i} S$ for some $i$ and some $U_ i$ in $(S_ i)_{affine, {\acute{e}tale}}$ and

\[ (\mathop{\mathrm{colim}}\nolimits _{i \in I} h_ i^{-1}R^ p g_{i, *} \mathcal{F}_ i)(U) = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} (R^ p g_{i', *}\mathcal{F}_{i'})(U_ i \times _{S_ i} S_{i'}) \]

by Sites, Lemma 7.18.4 and the construction of the transition maps in the system described above. Since $R^ pg_{i', *}\mathcal{F}_{i'}$ is the sheaf associated to the presheaf $U_{i'} \mapsto H^ p(U_{i'} \times _{S_{i'}} X_{i'}, \mathcal{F}_{i'})$ and since $R^ pg_*\mathcal{F}$ is the sheaf associated to the presheaf $U \mapsto H^ p(U \times _ S X, \mathcal{F})$ (Lemma 59.51.6) we obtain a canonical commutative diagram

\[ \xymatrix{ \mathop{\mathrm{colim}}\nolimits _{i' \geq i} H^ p(U_ i \times _{S_ i} X_{i'}, \mathcal{F}_{i'}) \ar[r] \ar[d] & \mathop{\mathrm{colim}}\nolimits _{i' \geq i} (R^ p g_{i', *}\mathcal{F}_{i'})(U_ i \times _{S_ i} S_{i'}) \ar[d] \\ H^ p(U \times _ S X, \mathcal{F}) \ar[r] & R^ pg_*\mathcal{F}(U) } \]

Observe that the left hand vertical arrow is an isomorphism by Theorem 59.51.3. We're trying to show that the right hand vertical arrow is an isomorphism. However, we already know that the source and target of this arrow are sheaves on $S_{affine, {\acute{e}tale}}$. Hence it suffices to show: (1) an element in the target, locally comes from an element in the source and (2) an element in the source which maps to zero in the target locally vanishes. Part (1) follows immediately from the above and the fact that the lower horizontal arrow comes from a map of presheaves which becomes an isomorphism after sheafification. For part (2), say $\xi \in \mathop{\mathrm{colim}}\nolimits _{i' \geq i} (R^ p g_{i', *}\mathcal{F}_{i'})(U_ i \times _{S_ i} S_{i'})$ is in the kernel. Choose an $i' \geq i$ and $\xi _{i'} \in (R^ p g_{i', *}\mathcal{F}_{i'})(U_ i \times _{S_ i} S_{i'})$ representing $\xi $. Choose a standard étale covering $\{ U_{i', k} \to U_ i \times _{S_ i} S_{i'}\} _{k = 1, \ldots , m}$ such that $\xi _{i'}|_{U_{i', k}}$ comes from $\xi _{i', k} \in H^ p(U_{i', k} \times _{S_{i'}} X_{i'}, \mathcal{F}_{i'})$. Since it is enough to prove that $\xi $ dies locally, we may replace $U$ by the members of the étale covering $\{ U_{i', k} \times _{S_{i'}} S \to U = U_ i \times _{S_ i} S\} $. After this replacement we see that $\xi $ is the image of an element $\xi '$ of the group $\mathop{\mathrm{colim}}\nolimits _{i' \geq i} H^ p(U_ i \times _{S_ i} X_{i'}, \mathcal{F}_{i'})$ in the diagram above. Since $\xi '$ maps to zero in $R^ pg_*\mathcal{F}(U)$ we can do another replacement and assume that $\xi '$ maps to zero in $H^ p(U \times _ S X, \mathcal{F})$. However, since the left vertical arrow is an isomorphism we then conclude $\xi ' = 0$ hence $\xi = 0$ as desired.
$\square$

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