**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 $i$th 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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