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His definition 8.1.1: Let $f:X\rightarrow Y$ be a morphism, with $Y$ non-singular of dimension $n.$ Let $p_X: X' \rightarrow X$, $p_Y: Y' \rightarrow Y$ be morphisms of schemes $X',Y'$ to $X$ and $Y$ respectively, and let $x \in A_kX', y \in A_lY'.$ Form the fiber square (notes: products without subscripts are over Spec $k$. $\gamma_f$ is the graph morphism of $f$.)

$$\begin{array}{rcl}X'\times_Y Y' & \to & X'\times Y' \\ \downarrow & & \downarrow {\scriptsize p_x \times p_y} \\ X & \stackrel{\gamma_f}{\to} & X\times Y\end{array}$$ And define $x \cdot_f y$ by $\gamma_f^!(x \times y).$

It seems to me that the $X$ here is completely unnecessary, and that the desired cycle class depends only on the morphism $f \circ p_X$: Form the commutative diagram $$\begin{array}{rcl}X'\times_Y Y' & \to & X'\times Y' \\ \downarrow & & \downarrow {\scriptsize p_x \times p_y} \\ X & \stackrel{\gamma_f}{\to} & X\times Y \\ \downarrow & & \downarrow \\ Y & \stackrel{\delta_Y}\to & Y \times Y\end{array}$$

Then, as $Y$ is smooth, both the second and the third row are regular imbeddings of relative dimension equal to the dimension of $Y$, so the excess intersection formula gives us $\gamma_f^!(x\times y) = \delta_Y^!(x\times y),$ and $\delta_Y^!(x \times y)$ doesn't depend on $X$ at all, only on the morphism $f \circ p_X.$

Am I doing something wrong? If not, why introduce the $X$ if it doesn't actually do anything?

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