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. $\square$
Comments (0)
There are also: