Lemma 48.16.3. In Situation 48.16.1 let f : X \to Y and g : Y \to Z be composable morphisms of \textit{FTS}_ S. Then there is a canonical isomorphism (g \circ f)^! \to f^! \circ g^!.
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 \overline{a} be the right adjoint of Lemma 48.3.1 for \overline{X} \to \overline{Y} and let \overline{b} be the right adjoint of Lemma 48.3.1 for \overline{Y} \to Z. Then \overline{a} \circ \overline{b} is the right adjoint of Lemma 48.3.1 for the composition \overline{X} \to Z. Hence g^! = i^* \circ \overline{b} and (g \circ f)^! = (X \to \overline{X})^* \circ \overline{a} \circ \overline{b}. Let U be the inverse image of Y in \overline{X} so that we get the commutative diagram
Let \overline{a}' be the right adjoint of Lemma 48.3.1 for U \to Y. Then f^! = j^* \circ \overline{a}'. We obtain
by (48.4.1.1) and we can use it to define
which is an isomorphism on objects of D_\mathit{QCoh}^+(\mathcal{O}_ Z) by Lemma 48.4.4. 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} with dense image 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} with dense image 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
and moreover the compactifications X \to \overline{X}_1 and Y \to \overline{Y}_2 have dense image. We use \overline{a}_ i, \overline{a}'_ i, \overline{c}, and \overline{c}' for the right adjoint of Lemma 48.3.1 for \overline{X}_ i \to \overline{Y}_ i, U_ i \to Y, \overline{X}_1 \to \overline{X}_2, and U_1 \to U_2. Each of the squares
is cartesian (see More on Flatness, Lemma 38.32.1 part (c) for A, D, E and recall that U_ i is the inverse image of Y by \overline{X}_ i \to \overline{Y}_ i for B, C) and hence gives rise to a base change map (48.4.1.1) as follows
Denote f_1^! = j_1^* \circ \overline{a}'_1, f_2^! = j_2^* \circ \overline{a}'_2, g_1^! = i_1^* \circ \overline{b}_1, g_2^! = i_2^* \circ \overline{b}_2, (g \circ f)_1^! = (j_1' \circ j_1)^* \circ \overline{a}_1 \circ \overline{b}_1, and (g \circ f)^!_2 = (j_2' \circ j_2)^* \circ \overline{a}_2 \circ \overline{b}_2. The construction given in the first paragraph of the proof and in Lemma 48.16.2 uses
\gamma _ C for the map (g \circ f)^!_1 \to f_1^! \circ g_1^!,
\gamma _ B for the map (g \circ f)^!_2 \to f_2^! \circ g_2^!,
\gamma _ A for the map f_1^! \to f_2^!,
\gamma _ D for the map g_1^! \to g_2^!, and
\gamma _ E for the map (g \circ f)^!_1 \to (g \circ f)^!_2.
We have to show that the diagram
is commutative. We will use Lemmas 48.5.1 and 48.5.2 and with (abuse of) notation as in Remark 48.5.3 (in particular dropping \star products with identity transformations from the notation). We can write \gamma _ E = \gamma _ A \circ \gamma _ F 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 48.5.3). Thus it suffices to prove that \gamma _ D \circ \gamma _ C = \gamma _ B \circ \gamma _ F. Since both of these are equal to the map (48.4.1.1) for the square
we conclude. \square
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