The Stacks project

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 ( 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 ( 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 ( for the square

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

we conclude. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0ATX. Beware of the difference between the letter 'O' and the digit '0'.