**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 an element $c^ A(\mathcal{E}) \in A(X)$.

As a first step, 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. Hence $A(X) = \prod A(X_ r)$. Thus it suffices to define $c^ A(\mathcal{E})$ when $\mathcal{E}$ has constant rank $r$. In this case let $p : P \to X$ be the projective bundle of $\mathcal{E}$. We can uniquely define elements $c_ i^ A(\mathcal{E}) \in A^ i(X)$ for $i \geq 0$ such that $c_0^ A(\mathcal{E}) = 1$ and the equation

45.12.1.1
\begin{equation} \label{weil-equation-chern-classes} \sum \nolimits _{i = 0}^ r (-1)^ i c_1(\mathcal{O}_ P(1))^ i \cup p^*c^ A_{r - i}(\mathcal{E}) = 0 \end{equation}

is true. As usual we set $c^ A(\mathcal{E}) = c_0^ A(\mathcal{E}) + c_1^ A(\mathcal{E}) + \ldots + c_ r^ A(\mathcal{E})$ in $A(X)$.

If $\mathcal{E}$ is invertible, then $c^ A(\mathcal{E}) = 1 + c_1^ A(\mathcal{L})$. This follows immediately from the construction above.

The elements $c_ i^ A(\mathcal{E})$ are in the center of $A(X)$. Namely, to prove this we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ be the corresponding projective bundle. if $a \in A(X)$ then $p^*a \cup (-1)^ r c_1(\mathcal{O}_ P(1))^ r = (-1)^ r c_1(\mathcal{O}_ P(1))^ r \cup p^*a$ and hence we must have the same for all the other terms in the expression defining $c_ i^ A(\mathcal{E})$ as well and we conclude.

If $f : X' \to X$ is a morphism of $\mathcal{C}$, then $f^*c_ i^ A(\mathcal{E}) = c_ i^ A(f^*\mathcal{E})$. Namely, to prove this we may assume $\mathcal{E}$ has constant rank $r$. Let $p : P \to X$ and $p' : P' \to X'$ be the projective bundles corresponding to $\mathcal{E}$ and $f^*\mathcal{E}$. The induced morphism $g : P' \to P$ is a morphism of $\mathcal{C}$. The pullback by $g$ of the equality defining $c_ i^ A(\mathcal{E})$ is the corresponding equation for $f^*\mathcal{E}$ and we conclude.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider a short exact sequence

\[ 0 \to \mathcal{L} \to \mathcal{E} \to \mathcal{F} \to 0 \]

of finite locally free $\mathcal{O}_ X$-modules with $\mathcal{L}$ invertible. Then

\[ c^ A(\mathcal{E}) = c^ A(\mathcal{L}) c^ A(\mathcal{F}) \]

Namely, by the construction of $c^ A_ i$ we may assume $\mathcal{E}$ has constant rank $r + 1$ and $\mathcal{F}$ has constant rank $r$. The inclusion

\[ i : P' = \mathbf{P}(\mathcal{F}) \longrightarrow \mathbf{P}(\mathcal{E}) = P \]

is a morphism of $\mathcal{C}$ and it is the zero scheme of a regular section of the invertible module $\mathcal{L}^{\otimes -1} \otimes \mathcal{O}_ P(1)$. The element

\[ \sum \nolimits _{i = 0}^ r (-1)^ i c_1^ A(\mathcal{O}_ P(1))^ i \cup p^*c^ A_ i(\mathcal{F}) \]

pulls back to zero on $P'$ by definition. Hence we see that

\[ \left(c_1^ A(\mathcal{O}_ P(1)) - c_1^ A(\mathcal{L})\right) \cup \left(\sum \nolimits _{i = 0}^ r (-1)^ i c_1^ A(\mathcal{O}_ P(1))^ i \cup p^*c^ A_ i(\mathcal{F})\right) = 0 \]

in $A^*(P)$ by assumption (5) on our cohomology $A$. By definition of $c_1^ A(\mathcal{E})$ this gives the desired equality.

Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Consider a short exact sequence

\[ 0 \to \mathcal{E} \to \mathcal{F} \to \mathcal{G} \to 0 \]

of finite locally free $\mathcal{O}_ X$-modules. Then

\[ c^ A(\mathcal{F}) = c^ A(\mathcal{E}) c^ A(\mathcal{G}) \]

Namely, by the construction of $c^ A_ i$ we may assume $\mathcal{E}$, $\mathcal{F}$, and $\mathcal{G}$ have constant ranks $r$, $s$, and $t$. We prove it by induction on $r$. The case $r = 1$ was done above. If $r > 1$, then it suffices to check this after pulling back by the morphism $\mathbf{P}(\mathcal{E}^\vee ) \to X$. Thus we may assume we have an invertible submodule $\mathcal{L} \subset \mathcal{E}$ such that both $\mathcal{E}' = \mathcal{E}/\mathcal{L}$ and $\mathcal{F}' = \mathcal{E}/\mathcal{L}$ are finite locally free (of ranks $s - 1$ and $t - 1$). Then we have

\[ c^ A(\mathcal{E}) = c^ A(\mathcal{L}) c^ A(\mathcal{E}') \quad \text{and}\quad c^ A(\mathcal{F}) = c^ A(\mathcal{L}) c^ A(\mathcal{F}') \]

Since we have the short exact sequence

\[ 0 \to \mathcal{E}' \to \mathcal{F}' \to \mathcal{G} \to 0 \]

we see by induction hypothesis that

\[ c^ A(\mathcal{F}') = c^ A(\mathcal{E}') c^ A(\mathcal{G}) \]

Thus the result follows from a formal calculation.

At this point for $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we can define $c^ A : K_0(\textit{Vect}(X)) \to A(X)$. Namely, we send a generator $[\mathcal{E}]$ to $c^ A(\mathcal{E})$ and we extend multiplicatively. Thus for example $c^ A(-[\mathcal{E}]) = c^ A(\mathcal{E})^{-1}$ is the formal inverse of $a^ A([\mathcal{E}])$. The multiplicativity in short exact sequences shown above guarantees that this works.

Uniqueness. Suppose $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and $\mathcal{E}$ is a finite locally free $\mathcal{O}_ X$-module. We want to show that conditions (1), (2), and (3) of the lemma uniquely determine $c^ A([\mathcal{E}])$. To prove this we may assume $\mathcal{E}$ has constant rank $r$; this already uses (2). Then we may use induction on $r$. If $r = 1$, then uniqueness follows from (3). If $r > 1$ we pullback using (2) to the projective bundle $p : P \to X$ and we see that we may assume we have a short exact sequence $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ with $\mathcal{E}'$ and $\mathcal{E}''$ having lower rank. By induction hypothesis $c^ A(\mathcal{E}')$ and $c^ A(\mathcal{E}'')$ are uniquely determined. Thus uniqueness for $\mathcal{E}$ by the axiom (1).
$\square$

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