The question is about the accepted answer here. I decided not to ask in a comment there since the original asker is no longer active.

In "Step 2" of the answer given there, Roland claims, "This is enough to work locally. In this case, we can assume that $F\simeq E \oplus L$, and that $E$ and $L$ are free with basis $e_{1},\ldots ,e_{n}$ and $l$."

I don't see at all why this is true. Obviously it is the case for $L$, since it is a locally free sheaf of rank $1$. But $E$ is just a coherent sheaf. Why should it be locally free? For convenience, let's just assume this is on a noetherian scheme.


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