TITLE: Sheaf with free stalks
QUESTION [9 upvotes]: Say we are given a complex manifold $X$ and an $\mathcal{O}_X$-module $\mathcal{F}$. Assume that for any point $P\in X$ the stalk $\mathcal{F}_P$ is a free $(\mathcal{O}_X)_P$-module of finite rank. Does it imply that $\mathcal{F}$ is locally free? If not, what do you need to know additionally about  $\mathcal{F}$ to make it true?
Note that if we were looking at the case of schemes then it would be wrong in general. Mathoverflow answer to a related question is 
here
Remark: As it was pointed out by Francesco Polizzi, this is true if $\mathcal{F}$ is coherent. What if we do not know it apriori?

REPLY [4 votes]: This is a small modification of Donu's answer.
Let $\mathcal F=(i_x)_\ast\mathcal O_x$ (the skyscraper sheaf of $\mathcal O_x$ over $x$) for some $x\in X$.  Then $\mathcal F$ is locally free of rank $1$ at $x$, and is locally free of rank $0$ everywhere else.  Clearly $\mathcal F$ is not locally free near $x$, since it doesn't even have locally constant rank.