Lemma 63.9.2. Let $f : X \to Y$ and $g : Y \to Z$ be separated morphisms of finite type of quasi-compact and quasi-separated schemes. Then there is a canonical isomorphism $Rg_! \circ Rf_! \to R(g \circ f)_!$.
Proof. Choose a compactification $i : Y \to \overline{Y}$ of $Y$ over $Z$. Choose a compactification $X \to \overline{X}$ of $X$ over $\overline{Y}$. This uses More on Flatness, Theorem 38.33.8 and Lemma 38.32.2 twice. Let $U$ be the inverse image of $Y$ in $\overline{X}$ so that we get the commutative diagram
Then we have
The first equality is the definition of $R(g \circ f)_!$. The second equality uses the identifications $R(\overline{g} \circ \overline{f})_* = R\overline{g}_* \circ R\overline{f}_*$ and $(j' \circ j)_! = j'_! \circ j_!$ of Lemma 63.3.13. The identification $i_! \circ Rf'_* \to R\overline{f}_* \circ j_!$ used in the third equality is Lemma 63.8.1. The final fourth equality is the definition of $Rg_!$ and $Rf_!$. To finish the proof we show that this isomorphism is independent of choices made.
Suppose we have two diagrams
We can first choose a compactification $i : Y \to \overline{Y}$ of $Y$ over $Z$ which dominates both $\overline{Y}_1$ and $\overline{Y}_2$, see More on Flatness, Lemma 38.32.1. By More on Flatness, Lemma 38.32.3 and Categories, Lemmas 4.27.13 and 4.27.14 we can choose a compactification $X \to \overline{X}$ of $X$ over $\overline{Y}$ with morphisms $\overline{X} \to \overline{X}_1$ and $\overline{X} \to \overline{X}_2$ and such that the composition $\overline{X} \to \overline{Y} \to \overline{Y}_1$ is equal to the composition $\overline{X} \to \overline{X}_1 \to \overline{Y}_1$ and such that the composition $\overline{X} \to \overline{Y} \to \overline{Y}_2$ is equal to the composition $\overline{X} \to \overline{X}_2 \to \overline{Y}_2$. Thus we see that it suffices to compare the maps determined by our diagrams when we have a commutative diagram as follows
Each of the squares
gives rise to an isomorphism as follows
by applying the map from Lemma 63.8.1 (which is the same as the map in Lemma 63.8.6 in case the left vertical arrow is the identity). Let us write
The construction given in the first paragraph of the proof and in Lemma 63.9.1 uses
$\gamma _ C$ for the map $G_1 \circ F_1 \to C_1$,
$\gamma _ B$ for the map $G_2 \circ F_2 \to C_2 $,
$\gamma _ A$ for the map $F_2 \to F_1$,
$\gamma _ D$ for the map $G_2 \to G_1$, and
$\gamma _ E$ for the map $C_2 \to C_1$.
This implies that we have to show that the diagram
is commutative. We will use Lemmas 63.8.2 and 63.8.3 and with (abuse of) notation as in Remark 63.8.4 (in particular dropping $\star $ products with identity transformations from the notation). We can write $\gamma _ E = \gamma _ F \circ \gamma _ A$ where
Thus we see that
the last equality because the two squares $A$ and $B$ only intersect in one point (similar to the last argument in Remark 63.8.4). Thus it suffices to prove that $\gamma _ C \circ \gamma _ D = \gamma _ F \circ \gamma _ B$. Since both of these are equal to the map for the square
we conclude. $\square$
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