Lemma 62.9.1. 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 62.8.6 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 62.8.6 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 62.8.6 is a special case of the map of Lemma 62.8.1 (as $j_1$ and $j_2$ are separated) this follows immediately from Lemma 62.8.2. $\square$

[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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