Proof.
Let X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) and let \mathcal{E} be a finite locally free \mathcal{O}_ X-module. We first show how to define the rank r^ A(\mathcal{E}) \in A^0(X). Namely, let X = \bigcup X_ r be the decomposition into open and closed subschemes such that \mathcal{E}|_{X_ r} has constant rank r. Since X is quasi-compact, this decomposition is finite, say X = X_0 \amalg X_1 \amalg \ldots \amalg X_ n. Then A(X) = A(X_0) \times A(X_1) \times \ldots \times A(X_ n). Thus we can define r^ A(\mathcal{E}) = (0, 1, \ldots , n) \in A^0(X).
Let P_ p(c_1, \ldots , c_ p) be the polynomials constructed in Chow Homology, Example 42.43.6. Then we can define
ch^ A(\mathcal{E}) = r^ A(\mathcal{E}) + \sum \nolimits _{i \geq 1} (1/i!) P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) \in \prod \nolimits _{i \geq 0} A^ i(X)
where ci^ A are the Chern classes of Proposition 45.12.1. It follows immediately that we have property (2) and (3) of the lemma.
We still have to show the following three statements
If 0 \to \mathcal{E}_1 \to \mathcal{E} \to \mathcal{E}_2 \to 0 is a short exact sequence of finite locally free \mathcal{O}_ X-modules on X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), then ch^ A(\mathcal{E}) = ch^ A(\mathcal{E}_1) + ch^ A(\mathcal{E}_2).
If \mathcal{E}_1 and \mathcal{E}_2 \to 0 are finite locally free \mathcal{O}_ X-modules on X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), then ch^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = ch^ A(\mathcal{E}_1) ch^ A(\mathcal{E}_2).
Namely, the first will prove that ch^ A factors through K_0(\textit{Vect}(X)) and the first and the second will combined show that ch^ A is a ring map.
To prove these statements we can reduce to the case where \mathcal{E}_1 and \mathcal{E}_2 have constant ranks r_1 and r_2. In this case the equalities in A^0(X) are immediate. To prove the equalities in higher degrees, by Lemma 45.12.2 we may assume that \mathcal{E}_1 and \mathcal{E}_2 have filtrations whose graded pieces are invertible modules \mathcal{L}_{1, j}, j = 1, \ldots , r_1 and \mathcal{L}_{2, j}, j = 1, \ldots , r_2. Using the multiplicativity of Chern classes we get
c_ i^ A(\mathcal{E}_1) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}))
where s_ i is the ith elementary symmetric function as in Chow Homology, Example 42.43.6. Similarly for c_ i^ A(\mathcal{E}_2). In case (1) we get
c_ i^ A(\mathcal{E}) = s_ i(c_1^ A(\mathcal{L}_{1, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}), c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{2, r_2}))
and for case (2) we get
c_ i^ A(\mathcal{E}_1 \otimes \mathcal{E}_2) = s_ i(c_1^ A(\mathcal{L}_{1, 1}) + c_1^ A(\mathcal{L}_{2, 1}), \ldots , c_1^ A(\mathcal{L}_{1, r_1}) + c_1^ A(\mathcal{L}_{2, r_2}))
By the definition of the polynomials P_ i we see that this means
P_ i(c^ A_1(\mathcal{E}_1), \ldots , c^ A_ i(\mathcal{E}_1)) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i
and similarly for \mathcal{E}_2. In case (1) we have also
P_ i(c^ A_1(\mathcal{E}), \ldots , c^ A_ i(\mathcal{E})) = \sum \nolimits _{j = 1, \ldots , r_1} c_1^ A(\mathcal{L}_{1, j})^ i + \sum \nolimits _{j = 1, \ldots , r_2} c_1^ A(\mathcal{L}_{2, j})^ i
In case (2) we get accordingly
P_ i(c^ A_1(\mathcal{E}_1 \otimes \mathcal{E}_2), \ldots , c^ A_ i(\mathcal{E}_1 \otimes \mathcal{E}_2)) = \sum \nolimits _{j = 1, \ldots , r_1} \sum \nolimits _{j' = 1, \ldots , r_2} (c_1^ A(\mathcal{L}_{1, j}) + c_1^ A(\mathcal{L}_{2, j'}))^ i
Thus the desired equalities are now consequences of elementary identities between symmetric polynomials.
We omit the proof of uniqueness.
\square
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