It is well known that a non-abelian free group is residually a finite simple group. Katz and Magnus proved, in fact, that non-abelian free groups are residually alternating and residually $PSL\_{2}$. S. J. Pride has some nice results along these lines as well. The best result that I know of is the theorem of Weigel that can be formulated as follows. If $\mathfrak{X}$ is a group-theoretic class containing an infinite set of pairwise non-isomorphic finite non-abelian simple groups, then every non-abelian free group is residually an $\mathfrak{X}$-group.




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My question is this:



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> Is a non-abelian free group fully residually a finite non-abelian simple group?
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It seems likely that the answer to such an obvious question is known, but I have not been able to find it in the literature.


I should probably add that I suspect we can probably replace "finite non-abelian simple" with "alternating", but I haven't yet given any thought to the other infinite series. I'd like to learn whether anything is known before spending more time on this.



Is a non-abelian free group fully residually a finite non-abelian simple group?
Yes! See my recent preprint [Alternating quotients of free groups](http://arxiv.org/abs/1005.0015).


I expect that what you want is well known, but I too couldn't find it in the literature. In fact, I prove the much stronger result that free groups are something like 'locally extended fully residually alternating'. Specifically:



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> Let $F$ be any free group of rank at least two, let $H$ be a finitely generated subgroup of infinite index in $F$ and let $\{g\_1,\ldots,g\_n\}$ be a finite subset of $F\smallsetminus H$. Then there is a surjection $f$ from $F$ to a finite alternating group such that $f(g\_i)$ is not in $f(H)$ for any $i$. 
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