Lemma 45.9.1. Assume given (D0), (D1), and (D3) satisfying (A). For $f : X \to Y$ a morphism of nonempty equidimensional smooth projective schemes over $k$ we have $f_*(f^*b \cup a) = b \cup f_*a$. If $g : Y \to Z$ is a second morphism with $Z$ nonempty smooth projective and equidimensional, then $g_* \circ f_* = (g \circ f)_*$.

**Proof.**
The first equality holds because

\[ \int _ Y c \cup b \cup f_*a = \int _ X f^*c \cup f^*b \cup a = \int _ Y c \cup f_*(f^*b \cup a). \]

The second equality holds because

\[ \int _ Z c \cup (g \circ f)_*a = \int _ X (g \circ f)^*c \cup a = \int _ X f^* g^* c \cup a = \int _ Y g^*c \cup f_*a = \int _ Z c \cup g_*f_*a \]

This ends the proof. $\square$

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