Lemma 80.30.1. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}$, $\mathcal{F}$ be finite locally free sheaves on $X$ of ranks $r$, $r - 1$ which fit into a short exact sequence

Then we have

in $A^*(X)$.

This section is the analogue of Chow Homology, Section 42.39.

Lemma 80.30.1. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}$, $\mathcal{F}$ be finite locally free sheaves on $X$ of ranks $r$, $r - 1$ which fit into a short exact sequence

\[ 0 \to \mathcal{O}_ X \to \mathcal{E} \to \mathcal{F} \to 0 \]

Then we have

\[ c_ r(\mathcal{E}) = 0, \quad c_ j(\mathcal{E}) = c_ j(\mathcal{F}), \quad j = 0, \ldots , r - 1 \]

in $A^*(X)$.

**Proof.**
The proof is identical to the proof of Chow Homology, Lemma 42.39.1 replacing the lemmas used there by Lemmas 80.26.9, 80.24.1, 80.19.4, and 80.28.1.
$\square$

Lemma 80.30.2. In Situation 80.2.1 let $X/B$ be good. Let $\mathcal{E}$, $\mathcal{F}$ be finite locally free sheaves on $X$ of ranks $r$, $r - 1$ which fit into a short exact sequence

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

where $\mathcal{L}$ is an invertible sheaf. Then

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

in $A^*(X)$.

**Proof.**
The proof is identical to the proof of Chow Homology, Lemma 42.39.2 replacing the lemmas used there by Lemmas 80.30.1 and 80.29.1.
$\square$

Lemma 80.30.3. In Situation 80.2.1 let $X/B$ be good. Suppose that $\mathcal{E}$ sits in an exact sequence

\[ 0 \to \mathcal{E}_1 \to \mathcal{E} \to \mathcal{E}_2 \to 0 \]

of finite locally free sheaves $\mathcal{E}_ i$ of rank $r_ i$. The total chern classes satisfy

\[ c(\mathcal{E}) = c(\mathcal{E}_1) c(\mathcal{E}_2) \]

in $A^*(X)$.

**Proof.**
The proof is identical to the proof of Chow Homology, Lemma 42.39.3 replacing the lemmas used there by Lemmas 80.26.9, 80.30.2, and 80.28.1.
$\square$

Lemma 80.30.4. In Situation 80.2.1 let $X/B$ be good. Let ${\mathcal L}_ i$, $i = 1, \ldots , r$ be invertible $\mathcal{O}_ X$-modules. Let $\mathcal{E}$ be a locally free rank $\mathcal{O}_ X$-module endowed with a filtration

\[ 0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E} \]

such that $\mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i$. Set $c_1({\mathcal L}_ i) = x_ i$. Then

\[ c(\mathcal{E}) = \prod \nolimits _{i = 1}^ r (1 + x_ i) \]

in $A^*(X)$.

**Proof.**
Apply Lemma 80.30.2 and induction.
$\square$

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