## 42.60 Gysin maps for diagonals

Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a smooth morphism of schemes locally of finite type over $S$. Then the diagonal morphism $\Delta : X \longrightarrow X \times _ Y X$ is a regular immersion, see More on Morphisms, Lemma 37.60.18. Thus we have the gysin map

$\Delta ^! \in A^*(X \to X \times _ Y X)^\wedge$

constructed in Section 42.59. If $X \to Y$ has constant relative dimension $d$, then $\Delta ^! \in A^ d(X \to X \times _ Y X)$.

Lemma 42.60.1. In the situation above we have $\Delta ^! \circ \text{pr}_ i^! = 1$ in $A^0(X)$.

Proof. Observe that the projections $\text{pr}_ i : X \times _ Y X \to X$ are smooth and hence we have gysin maps for these projections as well. Thus the lemma makes sense and is a special case of Lemma 42.59.6. $\square$

Proposition 42.60.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes locally of finite type over $S$. If $g$ is smooth of relative dimension $d$, then $A^ p(X \to Y) = A^{p - d}(X \to Z)$.

Proof. We will use that smooth morphisms are local complete intersection morphisms whose gysin maps exist (see Section 42.59). In particular we have $g^! \in A^{-d}(Y \to Z)$. Then we can send $c \in A^ p(X \to Y)$ to $c \circ g^! \in A^{p - d}(X \to Z)$.

Conversely, let $c' \in A^{p - d}(X \to Z)$. Denote $res(c')$ the restriction (Remark 42.33.5) of $c'$ by the morphism $Y \to Z$. Since the diagram

$\xymatrix{ X \times _ Z Y \ar[r]_{\text{pr}_2} \ar[d]_{\text{pr}_1} & Y \ar[d]^ g \\ X \ar[r]^ f & Z }$

is cartesian we find $res(c') \in A^{p - d}(X \times _ Z Y \to Y)$. Let $\Delta : Y \to Y \times _ Z Y$ be the diagonal and denote $res(\Delta ^!)$ the restriction of $\Delta ^!$ to $X \times _ Z Y$ by the morphism $X \times _ Z Y \to Y \times _ Z Y$. Since the diagram

$\xymatrix{ X \ar[r] \ar[d] & X \times _ Z Y \ar[d] \\ Y \ar[r]^-\Delta & Y \times _ Z Y }$

is cartesian we see that $res(\Delta ^!) \in A^ d(X \to X \times _ Z Y)$. Combining these two restrictions we obtain

$res(\Delta ^!) \circ res(c') \in A^ p(X \to Y)$

Thus we have produced maps $A^ p(X \to Y) \to A^{p - d}(X \to Z)$ and $A^{p - d}(X \to Z) \to A^ p(X \to Y)$. To finish the proof we will show these maps are mutually inverse.

Let us start with $c \in A^ p(X \to Y)$. Consider the diagram

$\xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ X \times _ Z Y \ar[r] \ar[d]^{\text{pr}_1} & Y \times _ Z Y \ar[r]_{p_2} \ar[d]^{p_1} & Y \ar[d]^ g \\ X \ar[r]^ f & Y \ar[r]^ g & Z }$

whose squares are carteisan. The lower two square of this diagram show that $res(c \circ g^!) = res(c) \cap p_2^!$ where in this formula $res(c)$ means the restriction of $c$ via $p_1$. Looking at the upper square of the diagram and using Lemma 42.59.7 we get $c \circ \Delta ^! = res(\Delta ^!) \circ res(c)$. We compute

\begin{align*} res(\Delta ^!) \circ res(c \circ g^!) & = res(\Delta ^!) \circ res(c) \circ p_2^! \\ & = c \circ \Delta ^! \circ p_2^! \\ & = c \end{align*}

The final equality by Lemma 42.60.1.

Conversely, let us start with $c' \in A^{p - d}(X \to Z)$. Looking at the lower rectangle of the diagram above we find $res(c') \circ g^! = \text{pr}_1^! \circ c'$. We compute

\begin{align*} res(\Delta ^!) \circ res(c') \circ g^! & = res(\Delta ^!) \circ \text{pr}_1^! \circ c' \\ & = c' \end{align*}

The final equality holds because the left two squares of the diagram show that $\text{id} = res(\Delta ^! \circ p_1^!) = res(\Delta ^!) \circ \text{pr}_1^!$. This finishes the proof. $\square$

## 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.

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