When does a sheaf exist with prescribed stalks?

I think the intended question is clear, but let me attempt to formulate it precisely:

If $X$ is a topological space, and $f$ is a function on $X$ such that $f(x)$ is an abelian group for every $x \in X$(I'll leave the codomain a mystery, it's a bit neater than phrasing this in terms of a functor), is there a sheaf $\mathcal{F}$ such that $\mathcal{F}_x = f(x)$?

On a Noetherian T1 space, (if we take $\mathcal{S}_x(A)$ to be the skyscraper sheaf with stalk $A$ at the point $x \in X$) taking $\mathcal{F} := \coprod_{x \in X} \mathcal{S}_x(f(x))$ should do the trick. Here the Noetherian property gives us that $\Gamma(U, \textrm{colim}_{\mathsf{Sh}} \, \mathcal{F}_i) \simeq \textrm{colim}_{\mathsf{Ab}} \,\, \Gamma(U, \mathcal{F}_i)$, and the T1 tells us that everything besides our copy of $f(x)$ is going to get killed off in the map from any section to the stalk(the detail here is that the coproduct is generated by the images of the canonical injections, so if we check that these all get killed off except the image of $f(x)$, we're almost done).

Can we improve these topological restrictions, or are these necessary? My first thought was to use a product (which wouldn't require the Noetherian condition), but the above argument doesn't work when we switch to the product.

EDIT: I fixed the title

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"Proscribed" means "forbidden". I think the word you want is "prescribed". – Zhen Lin May 15 '12 at 19:18
Also, obviously, your space $X$ has to be at least $T_0$ – otherwise you could have a pair of points which have the same neighbourhood filter, so any sheaf over $X$ will have isomorphic stalks over the two points. – Zhen Lin May 15 '12 at 19:45
Thank you Zhen, I was confused by the typical meaning of the prefix "pro-" – John Stalfos May 15 '12 at 20:43