Hartshorne's theorem II.7.1 claims that if $\phi : X\rightarrow \mathbf{P}^n_k$ is a morphism of a $k$-scheme into projective space, then $ \phi^*(\mathcal{O}(1)) $ is an invertible sheaf generated by global sections $s_i = \phi^*(x_i).$
Vakil's book, Theorem 16.4.1, claims that morphisms into projective space are in bijection with the data of (up to isomorphism) a line bundle on $X$ and $n+1$ global sections with no common zeros. This seems incompatible with Hartshorne's theorem--shouldn't Vakil require that that the $n+1$ global sections generate the line bundle?
Indeed, Vakil's proof claims that if $\phi$ is the morphism determined by the line bundle $\mathcal{L},$ then $\mathcal{L} \cong \phi^*\mathcal{O}(1).$ I do not think Vakil properly proves this if we do not know that $s_i = \phi^*(x_i)$$ generate the global sections. But let's assume Vakil's proof is right.
Wouldn't this immediately imply the line bundle is generated by global sections, since a sheaf and its pullback have isomorphic germs, isomorphic global sections, and the isomorphisms are compatible with the global sections to germs restrictions? It thus seems that Vakil proves a much stronger claim that any line bundle with $n+1$ global sections with no common zeroes is in fact generated by those global sections. This claim seems false.
Question: Is a morphism into projective space equivalent to the data of a line bundle and $n+1$ sections generating the bundle, or just $n+1$ sections with no common zeroes?
(It seems to me that this is what happens: given a line bundle and $n+1$ sections with no common zeroes, I can give a morphism into projective space. But the pullback of $\mathcal{O}(1)$ under that morphism is not always my original line bundle, but is instead some subbundle generated by the $n+1$ global sections.)