Lemma 48.16.2. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. The functor $f^!$ is, up to canonical isomorphism, independent of the choice of the compactification.

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

$j : X \to \overline{X},\quad \overline{f} : \overline{X} \to Y$

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

$j_1^* \circ \overline{c} \longrightarrow j_2^*$

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

$j_1^* \circ \overline{a}_1 \longrightarrow j_1^* \circ \overline{c} \circ \overline{a}_2 \longrightarrow j_2^* \circ \overline{a}_2$

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 compactififcatins. By More on Flatness, Lemma 38.32.1 part (c) we have $g_ i^{-1}(j_ i(X)) = j(X)$. Hence we get isomorpisms

$\alpha _{g_ i} : j^* \circ \overline{a} \longrightarrow j_ i^* \circ \overline{a}_ i$

by the previous paragraph. We obtain an isomorphism

$\alpha _{g_2} \circ \alpha _{g_1}^{-1} : j_1^* \circ \overline{a}_1 \to j_2^* \circ \overline{a}_2$

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

$\alpha _{g_ i} \circ \alpha _ h = \alpha _{g_ i \circ h} = \alpha _{g'_ i \circ h'} = \alpha _{g'_ i} \circ \alpha _{h'}$

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

$\alpha _ g \circ \alpha _ h = \alpha _{g \circ h} : j_1^* \circ \overline{a}_1 \to j_3^* \circ \overline{a}_3$

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

$j_1^* \circ \overline{c} \longrightarrow j_2^* \quad \text{and}\quad j_2^* \circ \overline{d} \longrightarrow j_3^*$

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

$j_1^* \circ \overline{e} \longrightarrow j_3^*$

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

$\alpha _ h \circ \alpha _ g : j_1^* \circ \overline{a}_1 \to j_1^* \circ \overline{c} \circ \overline{a}_2 \to j_2^* \circ \overline{a}_2 \to j_2^* \circ \overline{d} \circ \overline{a}_3 \to j_3^* \circ \overline{a}_3$

is the same as the composition

$\alpha _{h \circ g} : j_1^* \circ \overline{a}_1 \to j_1^* \circ \overline{e} \circ \overline{a}_3 \to j_3^* \circ \overline{a}_3$

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

$\xymatrix{ \overline{a}_1 \ar[r] \ar[d] & \overline{c} \circ \overline{a}_2 \ar[d] \\ \overline{e} \circ \overline{a}_3 \ar[r] & \overline{c} \circ \overline{d} \circ \overline{a}_3 }$

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

$\xymatrix{ R\overline{f}_{1, *} \ar[r] \ar[d] & Rg_* \circ R\overline{f}_{2, *} \ar[d] \\ R(h \circ g)_* \circ R\overline{f}_{3, *} \ar[r] & Rg_* \circ Rh_* \circ R\overline{f}_{3, *} }$

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

$j_1^* \circ \overline{c} \circ \overline{d} \to j_2^* \circ \overline{d} \to j_3^*$

is equal to the map

$j_1^* \circ \overline{e} \to j_3^*$

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 $\text{id}_ X$). $\square$

[1] This may fail with our definition of compactification. See More on Flatness, Section 38.32.

Comment #4670 by Bogdan on

It looks like the proof implicitly uses that $X=\overline{X_2}\times_{\overline{X_1}} X$ for any morphism of compactifications $\overline{X_2} \to \overline{X_1}$. But it seems false, if one uses definition from Tag 0ATT. For example, take $X=\overline{X_2}=\mathbf{P}^1_k$ and $\overline{X_1}=\mathbf{P}^1_k \sqcup \mathbf{P}^1_k$ with the evident projection map. I guess it should be a part of the definition of a morphism in the category of compactifications.

Comment #4671 by Bogdan on

Sorry, I confused $\overline{X_1}$ and $\overline{X_2}$ in the example. It should read $\overline{X_2}=\mathbf{P}^1_k \sqcup \mathbf{P}^1_k$ and $\overline{X_1}=\mathbf{P}^1_k$.

Comment #4675 by on

Yes, you are right! Somehow we never say for a compactification $\overline{X}$ of $X$ that the image of $X$ should be dense in $\overline{X}$. This is a very natural thing to ask for and if we impose it, then the problem you point out goes away. I'm not sure why we didn't ask for this. Weird! Let me think a bit more before fixing this. In any case thanks very much!

Comment #4678 by on

OK, thanks very much again. This is now fixed as you can see. I decided not to change the definition of a compactification. The changes I made are here. In the commit message there is a tiny bit more discussion of the error.

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