isomorphism of fibre functors

If $\mathfrak{C}$ is a $k$-linear rigid abelian tensor category with End(1)=$k$(strictly speaking is isomorphic to $k$ as a $k$-algebra), and $k=\bar{k}$, and if $\omega_1$ and $\omega_2$ are two fibre functors (i.e.exact faithful $k$-linear tensor functor to $\text{Vec}_{k}$), then can I say that $\omega_1\cong \omega_2$?

In Deligne's article "catégories tannakiennes" in The Grothendieck Festschrift Volume II, he explained a similar question in a more general context:

$\mathfrak{C}$ is a $k$-linear rigid abelian tensor category with End(1)=$k$, $S$ is a $k$-scheme. $\omega_1$ and $\omega_2$ are two fibre functors over $S$ (with values in the category of locally free sheaves of finite ranks over $S$). Then there is an fpqc covering $T\to S$ such that $\omega_1\cong \omega_2$ over $T$.

This is an easy corollary of the main theorem of Tannakian category (see remarques 1.13 in Deligne's article). From this point of view my question is equivalent to finding a $k$-rational point on some specific component of the representing groupoid. But since the representing groupoid is in general not of finite type, this might be difficult, but should still be possible since this is a component of a groupoid not an arbitrary affine scheme.

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To clarify, is the only difference between your situation and that of a Tannakian category is that the fiber functors are not necessarily faithful? What is your actual question? – Keerthi Madapusi Pera Oct 1 '11 at 14:46
– Angelo Oct 1 '11 at 14:51
@Keerthi Madapusi Pera, I don't know how important the faithfulness is , but I would think it is still true without faithfulness. In Deligne's article that I mentioned, in the definition of the fibre functor over arbitrary base scheme (1.9), he didn't require faithfulness. But this not my main point. I would still be happy to know the result if you put faithfulness. – Lei Oct 1 '11 at 15:18
The result follows formally from the result of Deligne that you stated, because any fpqc cover of $Spec k$ will admit a section (since $k=\overline{k}$. – Keerthi Madapusi Pera Oct 1 '11 at 15:29
I dont think so, if I take $K$ to be some non-trivial extension of $k$, this is an fpqc cover without a section. – Lei Oct 1 '11 at 15:37

The answer is yes. As already noted, it is true if the category is generated by a finite set of objects. Let $G$ be the group of tensor automorphisms of one of the fibre functors. Then $G$ is an affine group scheme and the isomorphisms from one fibre functor to the second form a torsor for $G$. The group $G$ is an inverse limit of algebraic groups in which the transition maps are surjective. Correspondingly, the torsor is an inverse limit of trivial torsors in which the transition maps are surjective. Therefore the torsor itself is trivial (i.e., it has $k$-point).
@anon, thank you very much for your answer. But I still could not understand. I translate your argument in categorical language(in my understanding): if you write our category $\mathfrak{C}$ as a direct limite of full sub tensor abelian categories generated by 1 object, if on each sub category the two fibre functors are isomorphic then they are isomorphic on the whole category. Here I think one needs a proof, since one does not know whether or not the local isomorphisms are unique. In fact they should not be, like in the topological cases one choose paths. – Lei Oct 2 '11 at 7:32
@anon, directly at your proof, I dont see where the $k$-point comes from, could you be more precise? – Lei Oct 2 '11 at 7:33