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

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 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

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

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