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If $X$ is a variety, and if $\mathcal{F}$, $\mathcal{G}$ are locally free sheaves of finite rank on $X$, does a surjection $\mathcal{F}\rightarrow \mathcal{G}$ induce a closed embedding $\mathbb{P}(\mathcal{G})\hookrightarrow \mathbb{P}(\mathcal{F})$ of the associated projective bundles?

Here I define $\mathbb{P}(\mathcal{G}) = \textbf{Proj}(\text{Sym}(\mathcal{G}))$ as in Hartshorne for instance. I understand that a surjection of graded rings $A\rightarrow B$ preserving degrees induces a closed embedding of the corresponding schemes $\text{Proj}(B)\hookrightarrow \text{Proj}(A)$. If the surjection $\mathcal{F}\rightarrow \mathcal{G}$ gives rise to a surjection $\text{Sym}(\mathcal{F})\rightarrow\text{Sym}(\mathcal{G})$ which is surjective on sections, then for each open subset $U\subseteq X$ we would have a closed embedding $$\text{Proj}(\text{Sym}(\mathcal{G})(U))\hookrightarrow \text{Proj}(\text{Sym}(\mathcal{F})(U)).$$ In this case I was thinking that these embeddings could be glued together to form an embedding $\mathbb{P}(\mathcal{G})\hookrightarrow \mathbb{P}(\mathcal{F})$. However the functor from sheaves to sections does not preserve surjective morphisms in general, and I cannot seem to find a way around this, even for locally free sheaves.

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  • $\begingroup$ Why do you think it's necessary to have a surjection on the level of global sections? Also, do you know the simplest case? As you probably know if you've read that section in Hartshorne, a section of the projective bundle $\mathbb{P}(E)$ comes from a quotient $E \to L \to 0$ where $L$ is locally free of rank $1$. And a section of a projective bundle is the same thing as a subbundle of (projective) rank $0$. $\endgroup$ – Tabes Bridges Jun 21 '17 at 16:56
  • $\begingroup$ @TabesBridges The main situation I had in mind is when a sheaf $\mathcal{G}$ on $X$ is globally generated, so is equipped with a surjection $\mathcal{O}_X^{m+1}\rightarrow \mathcal{G}$. This occurs in the proof of part (b) of prop. II.7.10 of Hartshorne, and there it is claimed in that specific case there is a closed embedding $\mathbb{P}(\mathcal{G})\hookrightarrow \mathbb{P}_X^m$. The proof seems to suggest using the ring map $\text{Sym}(\mathcal{O}_X^{m+1})(X)\rightarrow \text{Sym}(\mathcal{G})(X)$ to show this. However I cannot see how this will work unless the ring map is surjective. $\endgroup$ – catfish Jun 21 '17 at 20:05
  • $\begingroup$ @TabesBridges I think my confusion is really whether a surjective morphism of sheaves $\mathcal{F}\rightarrow \mathcal{G}$ where $\mathcal{F}$ is free need be surjective on global sections. If I understand correctly, the more general case in the question statement would follow from this by restricting to open sets where the locally free sheaves become free. $\endgroup$ – catfish Jun 21 '17 at 20:08
  • $\begingroup$ At least when $X$ is quasi projective, this issue can be easily resolved. Fix an ample line bundle $L$. Then, a surjection $F\to G$ is the same as a surjection $F\otimes L^n\to G\otimes L^n$ and if $n>>0$, your global section property is satisfied. $\endgroup$ – Mohan Jun 22 '17 at 17:30
  • $\begingroup$ What is missing in the argument above is that the question is local on the base $X$, so we may assume that $X$ is affine. In that case, a surjective map of sheaves gives a surjection on global sections. See II.5.6. This also follows from the vanishing of cohomology on an affine scheme. $\endgroup$ – Tony Blair's Witch Project Jul 5 '17 at 21:07

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