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 $A^ p(X \to Y) = A^{p - d}(X \to Z)$.

[Proposition 17.4.2, F]

**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

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

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

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

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

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

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. $\square$

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