The Stacks project

Proof. The statement means that if we write $R(g \circ f)_* = Rg_* \circ Rf_*$ and $R(g' \circ f')_* = Rg'_* \circ Rf'_*$, then the isomorphism $m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ Rf_* \circ k_!$ of the outer rectangle is equal to the composition

of the two maps of the squares in the diagram. To prove this choose a K-injective complex $\mathcal{J}^\bullet $ of $\Lambda $-modules on $X'_{\acute{e}tale}$ and a quasi-isomorphism $k_!\mathcal{J}^\bullet \to \mathcal{I}^\bullet $ to a K-injective complex $\mathcal{I}^\bullet $ of $\Lambda $-modules on $X_{\acute{e}tale}$. The proof of Lemma 63.8.1 shows that the canonical map

is a quasi-isomorphism and this quasi-isomorphism produces the second arrow on applying $Rg_*$. By Cohomology on Sites, Lemma 21.20.10 the complex $f_*\mathcal{I}^\bullet $, resp. $f'_*\mathcal{J}^\bullet $ is a K-injective complex of $\Lambda $-modules on $Y_{\acute{e}tale}$, resp. $Y'_{\acute{e}tale}$. (Using this is cheating and could be avoided.) In particular, the same reasoning gives that the canonical map

is a quasi-isomorphism and this quasi-isomorphism represents the first arrow. Finally, the proof of Lemma 63.8.1 show that $g_*l_!f'_!\mathcal{J}^\bullet $ represents $Rg_*(l_!f'_*\mathcal{J}^\bullet )$ because $f'_*\mathcal{J}^\bullet $ is K-injective. Hence $Rg_*(a) = g_*(a)$ and the composition $g_*(a) \circ b$ is the arrow of Lemma 63.8.1 for the rectangle. $\square$