Lemma 48.20.8. Let $(S, \omega _ S^\bullet )$ be as in Situation 48.20.1. Let $f : X \to Y$ be a proper morphism of schemes of finite type over $S$. Let $\omega _ X^\bullet $ and $\omega _ Y^\bullet $ be dualizing complexes normalized relative to $\omega _ S^\bullet $. Let $a$ be the right adjoint of Lemma 48.3.1 for $f$. Then there is a canonical isomorphism $a(\omega _ Y^\bullet ) = \omega _ X^\bullet $.

**Proof.**
Let $p : X \to S$ and $q : Y \to S$ be the structure morphisms. If $X$ and $Y$ are separated over $S$, then this follows from the fact that $\omega _ X^\bullet = p^!\omega _ S^\bullet $, $\omega _ Y^\bullet = q^!\omega _ S^\bullet $, $f^! = a$, and $f^! \circ q^! = p^!$ (Lemma 48.16.3). In the general case we first use Lemma 48.20.6 to reduce to the case $Y = S$. In this case $X$ and $Y$ are separated over $S$ and we've just seen the result.
$\square$

## 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Your email address will not be published.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.

## Comments (0)