Lemma 63.9.1. Let f : X \to Y be a finite type separated morphism of quasi-compact and quasi-separated schemes. The functors Rf_! constructed above are, up to canonical isomorphism, independent of the choice of the compactification.
Proof. We will prove this for the functor Rf_! : D(X_{\acute{e}tale}, \Lambda ) \to D(Y_{\acute{e}tale}, \Lambda ) when \Lambda is a torsion ring; the case of the functor Rf_! : D^+_{tors}(X_{\acute{e}tale}, \Lambda ) \to D^+_{tors}(Y_{\acute{e}tale}, \Lambda ) is proved in exactly the same way.
Consider the category of compactifications of X over Y, which is cofiltered according to More on Flatness, Theorem 38.33.8 and Lemmas 38.32.1 and 38.32.2. To every choice of a compactification
j : X \to \overline{X},\quad \overline{f} : \overline{X} \to Y
the construction above associates the functor R\overline{f}_* \circ j_! : D(X_{\acute{e}tale}, \Lambda ) \to D(Y_{\acute{e}tale}, \Lambda ). Let's be a little more explicit. Given a complex \mathcal{K}^\bullet of sheaves of \Lambda -modules on X_{\acute{e}tale}, we choose a quasi-isomorphism j_!\mathcal{K}^\bullet \to \mathcal{I}^\bullet into a K-injective complex of sheaves of \Lambda -modules on \overline{X}_{\acute{e}tale}. Then our functor sends \mathcal{K}^\bullet to \overline{f}_*\mathcal{I}^\bullet .
Suppose given a morphism g : \overline{X}_1 \to \overline{X}_2 between compactifications j_ i : X \to \overline{X}_ i over Y. Then we get an isomorphism
R\overline{f}_{2, *} \circ j_{2, !} = R\overline{f}_{2, *} \circ Rg_* \circ j_{1, !} = R\overline{f}_{1, *} \circ j_{1, !}
using Lemma 63.8.6 in the first equality.
To finish the proof, since the category of compactifications of X over Y is cofiltered, it suffices to show compositions of morphisms of compactifications of X over Y are turned into compositions of isomorphisms of functors1. To do this, suppose that j_3 : X \to \overline{X}_3 is a third compactification and that h : \overline{X}_2 \to \overline{X}_3 is a morphism of compactifications. Then we have to show that the composition
R\overline{f}_{3, *} \circ j_{3, !} = R\overline{f}_{3, *} \circ Rh_* \circ j_{2, !} = R\overline{f}_{2, *} \circ j_{2, !} = R\overline{f}_{2, *} \circ Rg_* \circ j_{1, !} = R\overline{f}_{1, *} \circ j_{1, !}
is equal to the isomorphism of functors constructed using simply j_3, g \circ h, and j_1. A calculation shows that it suffices to prove that the composition of the maps
j_{3, !} \to Rh_* \circ j_{2, !} \to Rh_* \circ Rg_* \circ j_{1, !}
of Lemma 63.8.6 agrees with the corresponding map j_{3, !} \to R(h \circ g)_* \circ j_{1, !} via the identification R(h \circ g)_* = Rh_* \circ Rg_*. Since the map of Lemma 63.8.6 is a special case of the map of Lemma 63.8.1 (as j_1 and j_2 are separated) this follows immediately from Lemma 63.8.2. \square
[1] Namely, if \alpha , \beta : F \to G are morphisms of functors and \gamma : G \to H is an isomorphism of functors such that \gamma \circ \alpha = \gamma \circ \beta , then we conclude \alpha = \beta .
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