Lemma 42.62.3. Let $k$ be a field. Let $f : X \to Y$ be a morphism of schemes smooth over $k$. Then the gysin map exists for $f$ and $f^!(\alpha \cdot \beta ) = f^!\alpha \cdot f^!\beta $.
Proof. Observe that $X \to X \times _ k Y$ is an immersion of $X$ into a scheme smooth over $Y$. Hence the gysin map exists for $f$ (Definition 42.59.4). To prove the formula we may decompose $X$ and $Y$ into their connected components, hence we may assume $X$ is smooth over $k$ and equidimensional of dimension $d$ and $Y$ is smooth over $k$ and equidimensional of dimension $e$. Observe that $f^![Y]_ e = [X]_ d$ (see for example Lemma 42.59.8). Write $\alpha = c \cap [Y]_ e$ and $\beta = c' \cap [Y]_ e$ and hence $\alpha \cdot \beta = c \cap c' \cap [Y]_ e$, see Lemma 42.62.2. By Lemma 42.59.7 we know that $f^!$ commutes with both $c$ and $c'$. Hence
\begin{align*} f^!(\alpha \cdot \beta ) & = f^!(c \cap c' \cap [Y]_ e) \\ & = c \cap c' \cap f^![Y]_ e \\ & = c \cap c' \cap [X]_ d \\ & = (c \cap [X]_ d) \cdot (c' \cap [X]_ d) \\ & = (c \cap f^![Y]_ e) \cdot (c' \cap f^![Y]_ e) \\ & = f^!(\alpha ) \cdot f^!(\beta ) \end{align*}
as desired where we have used Lemma 42.62.2 for $X$ as well.
An alternative proof can be given by proving that $(f \times f)^!(\alpha \times \beta ) = f^!\alpha \times f^!\beta $ and using Lemma 42.59.6. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: