The Stacks project

Lemma 48.17.9. In Situation 48.16.1 let $f : X \to Y$ be a morphism of $\textit{FTS}_ S$. Assume $f$ is perfect (e.g., flat). Then

  1. $f^!$ maps $D_{\textit{Coh}}^ b(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^ b(\mathcal{O}_ X)$,

  2. the map $\mu _{f, K} : Lf^*K \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y \to f^!K$ of Lemma 48.16.5 is an isomorphism for all $K \in D_\mathit{QCoh}^+(\mathcal{O}_ Y)$.

Proof. (A flat morphism of finite presentation is perfect, see More on Morphisms, Lemma 37.61.5.) We begin with a series of preliminary remarks.

  1. We already know that $f^!$ sends $D_{\textit{Coh}}^+(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}^+(\mathcal{O}_ X)$, see Lemma 48.17.6.

  2. If $f$ is an open immersion, then (a) and (b) are true because we can take $\overline{X} = Y$ in the construction of $f^!$ and $\mu _ f$. See also Lemma 48.17.1.

  3. If $f$ is a perfect proper morphism, then (b) is true by Lemma 48.13.3.

  4. If there exists an open covering $X = \bigcup U_ i$ and (a) is true for $U_ i \to Y$, then (a) is true for $X \to Y$. Same for (b). This holds because the construction of $f^!$ and $\mu _ f$ commutes with passing to open subschemes.

  5. If $g : Y \to Z$ is a second perfect morphism in $\textit{FTS}_ S$ and (b) holds for $f$ and $g$, then $f^!g^!\mathcal{O}_ Z = Lf^*g^!\mathcal{O}_ Z \otimes _{\mathcal{O}_ X}^\mathbf {L} f^!\mathcal{O}_ Y$ and (b) holds for $g \circ f$ by the commutative diagram of Lemma 48.16.5.

  6. If (a) and (b) hold for both $f$ and $g$, then (a) and (b) hold for $g \circ f$. Namely, then $f^!g^!\mathcal{O}_ Z$ is bounded above (by the previous point) and $L(g \circ f)^*$ has finite cohomological dimension and (a) follows from (b) which we saw above.

From these points we see it suffices to prove the result in case $X$ is affine. Choose an immersion $X \to \mathbf{A}^ n_ Y$ (Morphisms, Lemma 29.39.2) which we factor as $X \to U \to \mathbf{A}^ n_ Y \to Y$ where $X \to U$ is a closed immersion and $U \subset \mathbf{A}^ n_ Y$ is open. Note that $X \to U$ is a perfect closed immersion by More on Morphisms, Lemma 37.61.8. Thus it suffices to prove the lemma for a perfect closed immersion and for the projection $\mathbf{A}^ n_ Y \to Y$.

Let $f : X \to Y$ be a perfect closed immersion. We already know (b) holds. Let $K \in D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Then $f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ X, K)$ (Lemma 48.17.4) and $f_*f^!K = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$. Since $f$ is perfect, the complex $f_*\mathcal{O}_ X$ is perfect and hence $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(f_*\mathcal{O}_ X, K)$ is bounded above. This proves that (a) holds. Some details omitted.

Let $f : \mathbf{A}^ n_ Y \to Y$ be the projection. Then (a) holds by repeated application of Lemma 48.17.3. Finally, (b) is true because it holds for $\mathbf{P}^ n_ Y \to Y$ (flat and proper) and because $\mathbf{A}^ n_ Y \subset \mathbf{P}^ n_ Y$ is an open. $\square$


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 0B6U. Beware of the difference between the letter 'O' and the digit '0'.