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Let $S$ be a scheme, $\mathcal{E}$ a locally free $\mathcal{O}_S$-module of finite rank $r$ and $\pi:P=Proj(\mathcal{E})\to S$ the projective bundle. There is a canonical surjective map of sheaves $\psi:\pi^*\mathcal{E}\to \mathcal{O}(1)$.

Now for all $S$-schemes $f:X\to S$, there is a bijection (functorial in $X$) between the set of maps of $S$-schemes $X\to P$ and the set of equivalence classes of pairs $(\mathcal{L},\varphi)$ where $\mathcal{L}$ is an invertible sheaf on $X$ and $\varphi:f^*\mathcal{E}\to \mathcal{L}$ is a surjective map of sheaves. The equivalence relation is that $(\mathcal{L}',\varphi')\sim (\mathcal{L},\varphi)$ if and only if there exists an isomorphism $\tau:\mathcal{L}\to\mathcal{L}'$ such that $\varphi'=\tau\circ\varphi$.

For $X=P$ and $f=\mathrm{id}_P$, one gets a universal equivalence class which is supposed to be that of $(\mathcal{O}(1),\psi)$. So is $(\mathcal{O}(1),\psi)$, or only its equivalence class, canonical ?

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The pair $(O(1), \psi)$ is unique up to unique isomorphism, because your $\tau$ is unique: an automorphism of $\mathcal{L}$ dominated by the identity on $f^{\ast}E$ must be the identity. Put another way, the pairs being classified here have no nontrivial automorphisms, so this is a moduli problem for which nothing confusing about isomorphisms messes it up in the end. – BCnrd Sep 25 '10 at 8:06
Then if I understand correctly, the stack classifying pairs $(\mathcal{L},\varphi)$ has isomorphisms but no automorphisms, or in other words is equivalent to a functor, the functor represented by $P$. And the special fact here is that there is a distinguished element in each isomorphism class, namely the pullback from $P$ of $O(1)$. – Matthieu Romagny Sep 25 '10 at 14:35
A more elegant way to express the end of your comment is to say that the canonical representative is the quotient sheaf $f^{\ast}(E)/\ker(\varphi)$. – BCnrd Sep 25 '10 at 14:41
All right, the last formulation is better in that it doesn't involve the map to $P$. Thanks for your help ! – Matthieu Romagny Sep 26 '10 at 8:29
Alternatively, one could speak in terms of classifying hyperplane subbundles in $E$, and thereby bypass all issues about isomorphisms. – BCnrd Sep 26 '10 at 15:48

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