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. $\square$