The Stacks project

Proof. The category of compactifications of $X$ over $Y$ is defined in More on Flatness, Section 38.32. By More on Flatness, Theorem 38.33.8 and Lemma 38.32.2 it is nonempty. To every choice of a compactification

the construction above associates the functor $j^* \circ \overline{a} : D_\mathit{QCoh}^+(\mathcal{O}_ Y) \to D_\mathit{QCoh}^+(\mathcal{O}_ X)$ where $\overline{a}$ is the right adjoint of $R\overline{f}_*$ constructed in Lemma 48.3.1.

Suppose given a morphism $g : \overline{X}_1 \to \overline{X}_2$ between compactifications $j_ i : X \to \overline{X}_ i$ over $Y$ such that $g^{-1}(j_2(X)) = j_1(X)$¹. Let $\overline{c}$ be the right adjoint of Lemma 48.3.1 for $g$. Then $\overline{c} \circ \overline{a}_2 = \overline{a}_1$ because these functors are adjoint to $R\overline{f}_{2, *} \circ Rg_* = R(\overline{f}_2 \circ g)_*$. By (48.4.1.1) we have a canonical transformation

of functors $D^+_\mathit{QCoh}(\mathcal{O}_{\overline{X}_2}) \to D^+_\mathit{QCoh}(\mathcal{O}_ X)$ which is an isomorphism by Lemma 48.4.4. The composition

is an isomorphism of functors which we will denote by $\alpha _ g$.

Consider two compactifications $j_ i : X \to \overline{X}_ i$, $i = 1, 2$ of $X$ over $Y$. By More on Flatness, Lemma 38.32.1 part (b) we can find a compactification $j : X \to \overline{X}$ with dense image and morphisms $g_ i : \overline{X} \to \overline{X}_ i$ of compactifications. By More on Flatness, Lemma 38.32.1 part (c) we have $g_ i^{-1}(j_ i(X)) = j(X)$. Hence we get isomorpisms

by the previous paragraph. We obtain an isomorphism

To finish the proof we have to show that these isomorphisms are well defined. We claim it suffices to show the composition of isomorphisms constructed in the previous paragraph is another (for a precise statement see the next paragraph). We suggest the reader check this is true on a napkin, but we will also completely spell it out in the rest of this paragraph. Namely, consider a second choice of a compactification $j' : X \to \overline{X}'$ with dense image and morphisms of compactifications $g'_ i : \overline{X}' \to \overline{X}_ i$. By More on Flatness, Lemma 38.32.1 we can find a compactification $j'' : X \to \overline{X}''$ with dense image and morphisms of compactifications $h : \overline{X}'' \to \overline{X}$ and $h' : \overline{X}'' \to \overline{X}'$. We may even assume $g_1 \circ h = g'_1 \circ h'$ and $g_2 \circ h = g'_2 \circ h'$. The result of the next paragraph gives

for $i = 1, 2$. Since these are all isomorphisms of functors we conclude that $\alpha _{g_2} \circ \alpha _{g_1}^{-1} = \alpha _{g'_2} \circ \alpha _{g'_1}^{-1}$ as desired.

Suppose given compactifications $j_ i : X \to \overline{X}_ i$ for $i = 1, 2, 3$. Suppose given morphisms $g : \overline{X}_1 \to \overline{X}_2$ and $h : \overline{X}_2 \to \overline{X}_3$ of compactifications such that $g^{-1}(j_2(X)) = j_1(X)$ and $h^{-1}(j_2(X)) = j_3(X)$. Let $\overline{a}_ i$ be as above. The claim above means that

Let $\overline{c}$, resp. $\overline{d}$ be the right adjoint of Lemma 48.3.1 for $g$, resp. $h$. Then $\overline{c} \circ \overline{a}_2 = \overline{a}_1$ and $\overline{d} \circ \overline{a}_3 = \overline{a}_2$ and there are canonical transformations

of functors $D^+_\mathit{QCoh}(\mathcal{O}_{\overline{X}_2}) \to D^+_\mathit{QCoh}(\mathcal{O}_ X)$ and $D^+_\mathit{QCoh}(\mathcal{O}_{\overline{X}_3}) \to D^+_\mathit{QCoh}(\mathcal{O}_ X)$ for the same reasons as above. Denote $\overline{e}$ the right adjoint of Lemma 48.3.1 for $h \circ g$. There is a canonical transformation

of functors $D^+_\mathit{QCoh}(\mathcal{O}_{\overline{X}_3}) \to D^+_\mathit{QCoh}(\mathcal{O}_ X)$ given by (48.4.1.1). Spelling things out we have to show that the composition

is the same as the composition

We split this into two parts. The first is to show that the diagram

commutes where the lower horizontal arrow comes from the identification $\overline{e} = \overline{c} \circ \overline{d}$. This is true because the corresponding diagram of total direct image functors

is commutative (insert future reference here). The second part is to show that the composition

is equal to the map

via the identification $\overline{e} = \overline{c} \circ \overline{d}$. This was proven in Lemma 48.5.1 (note that in the current case the morphisms $f', g'$ of that lemma are equal to $\text{id}_ X$). $\square$

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

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

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. $\square$

Proof. To show this we have to prove given morphisms $f : X \to Y$, $g : Y \to Z$, $h : Z \to T$ that

is commutative (for the meaning of the $\gamma $'s, see below). To do this we choose a compactification $\overline{Z}$ of $Z$ over $T$, then a compactification $\overline{Y}$ of $Y$ over $\overline{Z}$, and then a compactification $\overline{X}$ of $X$ over $\overline{Y}$. This uses More on Flatness, Theorem 38.33.8 and Lemma 38.32.2. Let $W \subset \overline{Y}$ be the inverse image of $Z$ under $\overline{Y} \to \overline{Z}$ and let $U \subset V \subset \overline{X}$ be the inverse images of $Y \subset W$ under $\overline{X} \to \overline{Y}$. This produces the following diagram

Without introducing tons of notation but arguing exactly as in the proof of Lemma 48.16.3 we see that the maps in the first displayed diagram use the maps (48.4.1.1) for the rectangles $A + B$, $B + C$, $A$, and $C$ as indicated. Since by Lemmas 48.5.1 and 48.5.2 we have $\gamma _{A + B} = \gamma _ A \circ \gamma _ B$ and $\gamma _{B + C} = \gamma _ C \circ \gamma _ B$ we conclude that the desired equality holds provided $\gamma _ A \circ \gamma _ C = \gamma _ C \circ \gamma _ A$. This is true because the two squares $A$ and $C$ only intersect in one point (similar to the last argument in Remark 48.5.3). $\square$

Proof. If $f$ is proper, then $f^! = a$ and we can use (48.8.0.1) and if $g$ is also proper, then Lemma 48.8.4 proves the commutativity of the diagram (in greater generality).

Let us define the map $\mu _{f, K}$. Choose a compactification $j : X \to \overline{X}$ of $X$ over $Y$. Since $f^!$ is defined as $j^* \circ \overline{a}$ we obtain $\mu _{f, K}$ as the restriction of the map (48.8.0.1)

to $X$. To see this is independent of the choice of the compactification we argue as in the proof of Lemma 48.16.2. We urge the reader to read the proof of that lemma first.

Assume given a morphism $g : \overline{X}_1 \to \overline{X}_2$ between compactifications $j_ i : X \to \overline{X}_ i$ over $Y$ such that $g^{-1}(j_2(X)) = j_1(X)$. Denote $\overline{c}$ the right adjoint for pushforward of Lemma 48.3.1 for the morphism $g$. The maps

fit into the commutative diagram

by Lemma 48.8.4. By Lemma 48.8.3 the maps $\sigma $ and $\tau $ restrict to an isomorphism over $X$. In fact, we can say more. Recall that in the proof of Lemma 48.16.2 we used the map (48.4.1.1) $\gamma : j_1^* \circ \overline{c} \to j_2^*$ to construct our isomorphism $\alpha _ g : j_1^* \circ \overline{a}_1 \to j_2^* \circ \overline{a}_2$. Pulling back to map $\sigma $ by $j_1$ we obtain the identity map on $j_2^*\left(L\overline{f}_2^*K \otimes ^\mathbf {L} \overline{a}_2(\mathcal{O}_ Y)\right)$ if we identify $j_1^*\overline{c}(\mathcal{O}_{\overline{X}_2})$ with $\mathcal{O}_ X$ via $j_1^* \circ \overline{c} \to j_2^*$, see Lemma 48.8.2. Similarly, the map $\tau : Lg^*\overline{a}_2(\mathcal{O}_ Y) \otimes ^\mathbf {L} \overline{c}(\mathcal{O}_{\overline{X}_2}) \to \overline{a}_1(\mathcal{O}_ Y) = \overline{c}(\overline{a}_2(\mathcal{O}_ Y))$ pulls back to the identity map on $j_2^*\overline{a}_2(\mathcal{O}_ Y)$. We conclude that pulling back by $j_1$ and applying $\gamma $ wherever we can we obtain a commutative diagram

The commutativity of this diagram exactly tells us that the map $\mu _{f, K}$ constructed using the compactification $\overline{X}_1$ is the same as the map $\mu _{f, K}$ constructed using the compactification $\overline{X}_2$ via the identification $\alpha _ g$ used in the proof of Lemma 48.16.2. Some categorical arguments exactly as in the proof of Lemma 48.16.2 now show that $\mu _{f, K}$ is well defined (small detail omitted).

Having said this, the commutativity of the diagram in the statement of our lemma follows from the construction of the isomorphism $(g \circ f)^! \to f^! \circ g^!$ (first part of the proof of Lemma 48.16.3 using $\overline{X} \to \overline{Y} \to Z$) and the result of Lemma 48.8.4 for $\overline{X} \to \overline{Y} \to Z$. $\square$