The Stacks project

Proof. By Lemma 42.35.3 it suffices to show that if $X$ is integral then $c_ j(\mathcal{E}) \cap [X] = c_ j(\mathcal{F}) \cap [X]$. Let $(\pi : P \to X, \mathcal{O}_ P(1))$, resp. $(\pi ' : P' \to X, \mathcal{O}_{P'}(1))$ denote the projective space bundle associated to $\mathcal{E}$, resp. $\mathcal{F}$. The surjection $\mathcal{E} \to \mathcal{F}$ gives rise to a closed immersion

over $X$. Moreover, the element $1 \in \Gamma (X, \mathcal{O}_ X) \subset \Gamma (X, \mathcal{E})$ gives rise to a global section $s \in \Gamma (P, \mathcal{O}_ P(1))$ whose zero set is exactly $P'$. Hence $P'$ is an effective Cartier divisor on $P$ such that $\mathcal{O}_ P(P') \cong \mathcal{O}_ P(1)$. Hence we see that

for any cycle class $\alpha$ on $X$ by Lemma 42.31.1. By Lemma 42.38.2 we see that $\alpha _ j = c_ j(\mathcal{F}) \cap [X]$, $j = 0, \ldots , r - 1$ satisfy

Pushing this to $P$ and using the remark above as well as Lemma 42.26.4 we get

By the uniqueness of Lemma 42.38.2 we conclude that $c_ r(\mathcal{E}) \cap [X] = 0$ and $c_ j(\mathcal{E}) \cap [X] = \alpha _ j = c_ j(\mathcal{F}) \cap [X]$ for $j = 0, \ldots , r - 1$. Hence the lemma holds. $\square$

Proof. This relation really just says that $c_ i(\mathcal{E}) = c_ i(\mathcal{F}) + c_1(\mathcal{L})c_{i - 1}(\mathcal{F})$. By Lemma 42.40.1 we have $c_ j(\mathcal{E} \otimes \mathcal{L}^{\otimes -1}) = c_ j(\mathcal{F} \otimes \mathcal{L}^{\otimes -1})$ for $j = 0, \ldots , r$ were we set $c_ r(\mathcal{F} \otimes \mathcal{L}^{-1}) = 0$ by convention. Applying Lemma 42.39.1 we deduce

Setting $c_ i(\mathcal{E}) = c_ i(\mathcal{F}) + c_1(\mathcal{L})c_{i - 1}(\mathcal{F})$ gives a “solution” of this equation. The lemma follows if we show that this is the only possible solution. We omit the verification. $\square$

Proof. By Lemma 42.35.3 we may assume that $X$ is integral and we have to show the identity when capping against $[X]$. By induction on $r_1$. The case $r_1 = 1$ is Lemma 42.40.2. Assume $r_1 > 1$. Let $(\pi : P \to X, \mathcal{O}_ P(1))$ denote the projective space bundle associated to $\mathcal{E}_1$. Note that

1. $\pi ^* : \mathop{\mathrm{CH}}\nolimits _*(X) \to \mathop{\mathrm{CH}}\nolimits _*(P)$ is injective, and

2. $\pi ^*\mathcal{E}_1$ sits in a short exact sequence $0 \to \mathcal{F} \to \pi ^*\mathcal{E}_1 \to \mathcal{L} \to 0$ where $\mathcal{L}$ is invertible.

The first assertion follows from the projective space bundle formula and the second follows from the definition of a projective space bundle. (In fact $\mathcal{L} = \mathcal{O}_ P(1)$.) Let $Q = \pi ^*\mathcal{E}/\mathcal{F}$, which sits in an exact sequence $0 \to \mathcal{L} \to Q \to \pi ^*\mathcal{E}_2 \to 0$. By induction we have

Since $[P] = \pi ^*[X]$ we win by Lemma 42.38.5. $\square$

Proof. Apply Lemma 42.40.2 and induction. $\square$