[FOM] Fraenkel-Mostowski-Specker method and category theory
ablass at umich.edu
Tue Apr 4 09:48:20 EDT 2006
In his message of April 3, Solovay listed two situations in which
changing a group-and-filter pair (G,Gamma) leaves the resulting
Fraenkel-Mostowski-Specker topos C(G,Gamma) unchanged up to
equivalence of categories, namely:
> (a) If there is an isomorphism of G_1 with G_2 that carries
> Gamma_1 onto Gamma_2 then the two categories are equivalent.
> (b) Let G be a group and Gamma a normal filter of subgroups of
> G. Let H be a subgroup of G lying in Gamma, and let Gamma' be the
> of all subgroups of H lying in Gamma. Then C(G, Gamma) and C
> (H,Gamma') are equivalent.
I believe there are two more situations of this sort.
(c) Let N be the intersection of all the groups in the filter Gamma.
Then N is a normal subgroup of G, and the category C(G,Gamma) is
equivalent to C(G/N,Gamma'), where Gamma' consists of the groups H/N
for H in Gamma.
(d) Suppose G' is a subgroup of G with the property that, for each g
in G and each H in Gamma, the coset gH meets G'. (If we make G a
topological group by taking Gamma as a base of neighborhoods of the
identity, then this property is just that G' is dense in G.) Then C
(G,Gamma) is equivalent to C(G',Gamma'), where Gamma' consists of the
groups (H intersect G') for H in Gamma.
Of course, these observations don't address the "real" issue behind
Solovay's question, namely whether two categories of the form C
(G,Gamma) can be equivalent for "interesting" reasons.
By the way, situations (c) and (d) could be eliminated by insisting
that G be a complete, Hausdorff group with respect to the topology
mentioned in (d). According to (c) and (d), we can always replace G
by the completion of its universal Hausdorff quotient, so restricting
to complete, Hausdorff groups involves no loss of generality.
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