Trivial line bundle surjects onto any line bundle

Question

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.

Answer 1

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.