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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 37.62.18. 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.reference 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


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