1
$\begingroup$

My question is simple:

Is the category of profinite groups an accessible category?

Thank you

Edit: I will add the (hopefully simpler) question:

Is the category of profinite groups complete and does it admit a cogenerator?

$\endgroup$
  • $\begingroup$ Hint for the completeness: The forgetful functor from the category of profinite groups to the category of groups has a left-adjoint (pro-finite completion), so it preserves limits. That means if the category of profinte groups has some limits, the underlying group of the limit is the limit of the underlying groups in the category of groups. $\endgroup$ – archipelago Aug 17 '13 at 15:35
  • 1
    $\begingroup$ Now you can search for products and pullbacks in the category of profinite groups by taking them in the category of groups and thinking about a suitable topology. This will lead yout to the completeness of the category of profinite groups. $\endgroup$ – archipelago Aug 17 '13 at 15:37
  • $\begingroup$ Duplicate: math.stackexchange.com/questions/469677/… $\endgroup$ – Qiaochu Yuan Aug 17 '13 at 17:10
2
$\begingroup$

The category of profinite groups does admit a cogenerator. Just take the product over all finite groups with the product topology. If $f, g : G \to H$ is a parallel pair of morphisms between profinite groups, then $f, g$ can be distinguished by a map out of $H$ into some finite group. This finite group in turn embeds into the product over all finite groups.

$\endgroup$
2
$\begingroup$

This is a partial answere. Let $\mathcal{C}$ this category.

1) It's dual is finitely presentable: form profinite group this category is $Pro(Gr_f)$ where $Gr_f$ is the category of finite groups, for definition $Pro(Gr_f)=(Ind(Gr_f^{op}))^{op}$. Now, $Gr_f^{op}$ has finite colimits, then $Ind(Gr_f^{op})$ is equivalent to the category $Cont_f(Gr_f^{op}, Set)$ of functors $P: Gr_f^{op} \to Set$ such that $P^{op}$ preserving finite colimit, from T.1.46 of LPAC follow that $\mathcal{C}$ is finitely presentable.

2) From T.1.67 of LPAC follow that $\mathcal{C}$ cannot be locally presentable (isn't a complete lattice)

3) From c.2.47 of LPAC follow that $\mathcal{C}$ cannot be locally accessible and complete.

EDIT:

4) Now $Ind(Gr_f^{op})$ as any Ind-category (that is a finite cocompletion) has finite colimit, then $\mathcal{C}$ has finite limits. Furthermore $\mathcal{C}$ has (small) products ( as category of Hausdorff, compact, and totally disconnected, grops). THen $\mathcal{C}$ is complete.

THen from (3) $\mathcal{C}$ isn't locally accessible.

LPAC: LOCALLY PRESENTABLE AND ACCESSIBLE CATEGORIES. J.Adámek and J.Rosický.

$\endgroup$
1
$\begingroup$

The category of profinite groups is complete and admits a small strong cogenerating family because its opposite is locally finitely presentable; indeed:

  1. Any locally finitely presentable category is cocomplete and admits a small strong generating family.
  2. The category of profinite groups is the pro-completion of the category of finite groups, which means its opposite is the ind-completion of the opposite of the category of finite groups; the latter has finite colimits, and as a general fact the ind-completion of a small category with finite colimits is a locally finitely presentable category.

Because the category of profinite groups is pointed, this is enough to imply that it has a cogenerator: just take the product of the small cogenerating family.

Incidentally, the argument I suggested in the comments here goes through now: the opposite of a locally finitely presentable category is locally finitely presentable if and only if it is a preorder; but a complete category is accessible if and only if it is locally presentable. These facts can be found in [Adámek and Rosický, Locally presentable and accessible categories], as cited in Buschi Sergio's answer. The fact that the category of profinite groups is the pro-completion of the category of finite groups is actually non-trivial and is shown in [Johnstone, Stone spaces].

$\endgroup$
  • $\begingroup$ Can you indicate why it is nontrivial that the category of profinite groups is the pro-completion of the category of finite groups? What are the main steps here? $\endgroup$ – Martin Brandenburg Aug 17 '13 at 18:21
  • $\begingroup$ Well, it's trivial if you define profinite groups to be pro-objects in the category of finite groups. But the typical definition seems to be "topological group that has an underlying totally disconnected compact Hausdorff space". So one has to show that every such is indeed the inverse limit of an inverse system of finite groups, and that finite groups are "co-compact". That this is non-trivial is exemplified by the fact that the category of pro-discrete groups isn't just a subcategory of the category of topological groups. $\endgroup$ – Zhen Lin Aug 17 '13 at 18:33
  • $\begingroup$ See Johnstone "Stone Space" corollary pag.237, where the phrase "..expressible as limits .." can be declined as "..expressible as directed limits .." (because these limits come from Ind-object see T.18. in the cited book) $\endgroup$ – Buschi Sergio Aug 17 '13 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy