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

$\xymatrix{ X \ar[r]_ j \ar[d] & U \ar[dl] \ar[r]_{j'} & \overline{X} \ar[dl] \\ Y \ar[r]_ i \ar[d] & \overline{Y} \ar[dl] \\ Z }$

Let $\overline{a}'$ be the right adjoint of Lemma 48.3.1 for $U \to Y$. Then $f^! = j^* \circ \overline{a}'$. We obtain

$\gamma : (j')^* \circ \overline{a} \to \overline{a}' \circ i^*$

by (48.4.1.1) and we can use it to define

$(g \circ f)^! = (j' \circ j)^* \circ \overline{a} \circ \overline{b} = j^* \circ (j')^* \circ \overline{a} \circ \overline{b} \to j^* \circ \overline{a}' \circ i^* \circ \overline{b} = f^! \circ g^!$

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

$\vcenter { \xymatrix{ X \ar[r]_{j_1} \ar[d] & U_1 \ar[dl] \ar[r]_{j'_1} & \overline{X}_1 \ar[dl] \\ Y \ar[r]_{i_1} \ar[d] & \overline{Y}_1 \ar[dl] \\ Z } } \quad \text{and}\quad \vcenter { \xymatrix{ X \ar[r]_{j_2} \ar[d] & U_2 \ar[dl] \ar[r]_{j'_2} & \overline{X}_2 \ar[dl] \\ Y \ar[r]_{i_2} \ar[d] & \overline{Y}_2 \ar[dl] \\ Z } }$

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

$\xymatrix{ X \ar[rr]_{j_1} \ar@{=}[d] & & U_1 \ar[d] \ar[ddll] \ar[rr]_{j'_1} & & \overline{X}_1 \ar[d] \ar[ddll] \\ X \ar '[r][rr]^-{j_2} \ar[d] & & U_2 \ar '[dl][ddll] \ar '[r][rr]^-{j'_2} & & \overline{X}_2 \ar[ddll] \\ Y \ar[rr]^{i_1} \ar@{=}[d] & & \overline{Y}_1 \ar[d] \\ Y \ar[rr]^{i_2} \ar[d] & & \overline{Y}_2 \ar[dll] \\ Z }$

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

$\xymatrix{ X \ar[r] \ar[d] \ar@{}[dr]|A & U_1 \ar[d] \\ X \ar[r] & U_2 } \quad \xymatrix{ U_2 \ar[r] \ar[d] \ar@{}[dr]|B & \overline{X}_2 \ar[d] \\ Y \ar[r] & \overline{Y}_2 } \quad \xymatrix{ U_1 \ar[r] \ar[d] \ar@{}[dr]|C & \overline{X}_1 \ar[d] \\ Y \ar[r] & \overline{Y}_1 } \quad \xymatrix{ Y \ar[r] \ar[d] \ar@{}[dr]|D & \overline{Y}_1 \ar[d] \\ Y \ar[r] & \overline{Y}_2 } \quad \xymatrix{ X \ar[r] \ar[d] \ar@{}[dr]|E & \overline{X}_1 \ar[d] \\ X \ar[r] & \overline{X}_2 }$

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

$\begin{matrix} \gamma _ A : j_1^* \circ \overline{c}' \to j_2^* & \gamma _ B : (j_2')^* \circ \overline{a}_2 \to \overline{a}'_2 \circ i_2^* & \gamma _ C : (j_1')^* \circ \overline{a}_1 \to \overline{a}'_1 \circ i_1^* \\ \gamma _ D : i_1^* \circ \overline{d} \to i_2^* & \gamma _ E : (j'_1 \circ j_1)^* \circ \overline{c} \to (j'_2 \circ j_2)^* \end{matrix}$

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

1. $\gamma _ C$ for the map $(g \circ f)^!_1 \to f_1^! \circ g_1^!$,

2. $\gamma _ B$ for the map $(g \circ f)^!_2 \to f_2^! \circ g_2^!$,

3. $\gamma _ A$ for the map $f_1^! \to f_2^!$,

4. $\gamma _ D$ for the map $g_1^! \to g_2^!$, and

5. $\gamma _ E$ for the map $(g \circ f)^!_1 \to (g \circ f)^!_2$.

We have to show that the diagram

$\xymatrix{ (g \circ f)^!_1 \ar[r]_{\gamma _ E} \ar[d]_{\gamma _ C} & (g \circ f)^!_2 \ar[d]_{\gamma _ B} \\ f_1^! \circ g_1^! \ar[r]^{\gamma _ A \circ \gamma _ D} & f_2^! \circ g_2^! }$

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

$\xymatrix{ U_1 \ar[r] \ar[d] \ar@{}[rd]|F & \overline{X}_1 \ar[d] \\ U_2 \ar[r] & \overline{X}_2 }$

Thus we see that

$\gamma _ B \circ \gamma _ E = \gamma _ B \circ \gamma _ A \circ \gamma _ F = \gamma _ A \circ \gamma _ B \circ \gamma _ F$

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

$\xymatrix{ U_1 \ar[r] \ar[d] & \overline{X}_1 \ar[d] \\ Y \ar[r] & \overline{Y}_2 }$

we conclude. $\square$

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