# Trivial line bundle surjects onto any line bundle

Suppose $$S$$ is any scheme over an algebraically closed field $$k$$. Let $$\mathcal{O}_S$$ be the structure sheaf of $$S$$. Let $$L$$ be a line bundle on $$S$$, i.e., it is a locally free sheaf of rank $$1$$ on $$S$$. Is there always a way to write a surjective map of vector bundles $$\mathcal{O}_S\to L$$, where $$\mathcal{O}_S$$ is the trivial vector bundle of rank $$1$$?

In my mind, such a thing should always be geometrically possible, by "twisting" the copy of $$\mathcal{O}_S$$ to match with $$L$$. Of course, such a map would not be an isomorphism in general, unless $$L$$ was also trivial. I think either I am speaking nonsense or what I want is true for obvious reasons. Any remarks are helpful.

• KReiser has already answered your question, but in general I think what you are looking for is the notion of a globally generated sheaf, i.e. one which admits a surjection from a trivial vector bundle of some rank. Aug 25, 2021 at 3:01

No, there are many counterexamples. Any surjective map of line bundles on a locally ringed space is an isomorphism, because any surjective module endomorphism of the regular module over a local ring is in fact an isomorphism (if $$f:R\to R$$ is the endomorphism, then $$f(r)=r\cdot f(1)$$, and if $$f(u)=1$$, then $$uf(1)=1$$ so $$f$$ is multiplication by a unit). Therefore if $$S$$ has any line bundles not isomorphic to $$\mathcal{O}_S$$, you have found a counterexample.