Lemma 45.12.2. In the situation above. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $\mathcal{E}_ i$ be a finite collection of locally free $\mathcal{O}_ X$-modules of rank $r_ i$. There exists a morphism $p : P \to X$ in $\mathcal{C}$ such that
$p^* : A(X) \to A(P)$ is injective,
each $p^*\mathcal{E}_ i$ has a filtration whose successive quotients $\mathcal{L}_{i, 1}, \ldots , \mathcal{L}_{i, r_ i}$ are invertible $\mathcal{O}_ P$-modules.
Proof. We may assume $r_ i \geq 1$ for all $i$. We will prove the lemma by induction on $\sum (r_ i - 1)$. If this integer is $0$, then $\mathcal{E}_ i$ is invertible for all $i$ and we conclude by taking $\pi = \text{id}_ X$. If not, then we can pick an $i$ such that $r_ i > 1$ and consider the projective bundle $p : P \to X$ associated to $\mathcal{E}_ i$. We have a short exact sequence
of finite locally free $\mathcal{O}_ P$-modules of ranks $r_ i - 1$, $r_ i$, and $1$. Observe that $p^* : A(X) \to A(P)$ is injective by assumption. By the induction hypothesis applied to the finite locally free $\mathcal{O}_ P$-modules $\mathcal{F}$ and $p^*\mathcal{E}_{i'}$ for $i' \not= i$, we find a morphism $p' : P' \to P$ with properties stated as in the lemma. Then the composition $p \circ p' : P' \to X$ does the job. $\square$
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