Lemma 63.8.2. Consider a commutative diagram of schemes
\xymatrix{ X' \ar[r]_ k \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]_ l \ar[d]_{g'} & Y \ar[d]^ g \\ Z' \ar[r]^ m & Z }
with f, f', g and g' proper and k, l, and m separated and locally quasi-finite. Then the isomorphisms of Lemma 63.8.1 for the two squares compose to give the isomorphism for the outer rectangle (see proof for a precise statement).
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
m_! \circ Rg'_* \circ Rf'_* \to Rg_* \circ l_! \circ Rf'_* \to Rg_* \circ Rf_* \circ k_!
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
a : l_!f'_*\mathcal{J}^\bullet \to f_*\mathcal{I}^\bullet
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
b : m_!g'_*f'_*\mathcal{J}^\bullet \to g_*f_*\mathcal{I}^\bullet
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
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