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Let $\mathit{Profinite}_{\mathrm{Ab}}$ be the category of profinite abelian groups, and let $\mathit{Profinite}_{\mathrm{Set}}$ be the category of profinite sets. Does the forgetful functor

$$\mathit{Profinite}_{\mathrm{Ab}} \to \mathit{Profinite}_{Sets}$$

admit a left adjoint?

I am a beginner to this kind of question; I do not even know if both the domain and codomain categories admit all small colimits.

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  • $\begingroup$ Changed using "mathit", which corrects "$Profinite$" to "$\mathit{Profinite}$". $\endgroup$
    – YCor
    Commented Dec 24, 2019 at 23:57
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    $\begingroup$ Just using the universal properties, one sees that the profinite completion $G_I$ of the free abelian group $\mathrm{Z}^{(I)}$ is naturally the free profinite abelian group over the set $I$, in the sense that every map $I\to H$ for $H$ a profinite abelian group uniquely extends to a continuous homomorphism $G_I\to H$. (I see this is not the asked question, this is just a remark.) $\endgroup$
    – YCor
    Commented Dec 24, 2019 at 23:59
  • $\begingroup$ To adapt to the question, I guess that $I$ being profinite, one can start from $\mathbf{Z}^{(I)}$ and call a congruence subgroup, a subgroup $L$ of finite index such that the composed map $I\to \mathbf{Z}^{(I)}/L$ factors through some finite quotient of the profinite set $I$. Then the pro-congruence completion of $\mathbf{Z}^{(I)}$ (which is a quotient of the profinite completion) ought to be the required free profinite abelian group over the profinite set $I$. $\endgroup$
    – YCor
    Commented Dec 25, 2019 at 0:06
  • $\begingroup$ I think for some profinite set $S = \varprojlim S_i$, you can take $\varprojlim_i \widehat{\mathbb Z}[S_i]$. $\endgroup$
    – Julian Quast
    Commented Feb 11, 2021 at 12:15

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Yes, the free functor (i.e. the left adjoint to the "forgetful" functor) exists.

Let $Ab^{fin}$ be the category of finite abelian groups and $Set^{fin}$ the category of finite sets. Because each of these categories are essentially small and have finite limits, the categories of pro-objects $Pro(Ab^{fin})$ and $Pro(Set^{fin})$ are complete and cocomplete. In fact, they are opposite to locally finitely presentable categories -- if $C$ is essentially small with finite limits then $Pro(C) \simeq Fun^{lex}(C,Set)^{op}$ where $Fun^{lex}$ denotes the category of finite-limit-preserving functors. So equivalently, we're asking whether the forgetful functor $Fun^{lex}(Ab^{fin}, Set) \to Fun^{lex}(Set^{fin}, Set)$ has a right adjoint.

The forgetful functor $U_!: Fun^{lex}(Ab^{fin}, Set) \to Fun^{lex}(Set^{fin}, Set)$ is by definition the left Kan extension $Lan_{y_{Ab^{fin}}} (y_{Set^{fin}}\circ U^{op})$ where $y_C: C^{op} \to Fun^{lex}(C,Set)$ is the co-Yoneda embedding and $U: Ab^{fin} \to Set^{fin}$ is the forgetful functor. If we look at the full presheaf categories $Fun(C,Set)$ rather than $Fun^{lex}(C,Set)$, the right adjoint to this exists and is given by the functor $U^\ast: Fun(Set^{fin},Set) \to Fun(Ab^{fin}, Set)$ given by precomposition by the forgetful functor $U: Ab^{fin} \to Set^{fin}$.

Since $U: Ab^{fin} \to Set^{fin}$ preserves finite limits, $U^\ast$ restricts to a functor $Fun^{lex}(Set^{fin}, Set) \to Fun^{lex}(Ab^{fin}, Set)$, and because $Fun^{lex}$ is a full subcategory of $Fun$, the same calculation as at the link above shows that this functor is the adjoint we are searching for.

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  • $\begingroup$ Why does Fun^{lex} admit all small colimits? I understand it admits filtered colimits (which commute with finite limits in Set) but how do I compute a finite colimit in Fun^{lex}? $\endgroup$
    – Profinite Questioner
    Commented Dec 24, 2019 at 23:09
  • $\begingroup$ @ProfiniteQuestioner It is a reflective category of the presheaf category. $\endgroup$
    – Martin Brandenburg
    Commented Dec 25, 2019 at 11:52
  • $\begingroup$ I am missing something basic here. Do you conclude the existence of the left adjoint (to the inclusion of Fun^{lex} into Fun) by (i) constructing it (if so, how?), or by (ii) proving that Fun^{lex} admits all small colimits (if so, how?) and invoking the adjoint functor theorem, or (iii) exhibiting Fun^{lex} of a localization of Fun along certain morphisms (if so, how?). I am sorry these are such basic questions. $\endgroup$
    – Profinite Questioner
    Commented Dec 25, 2019 at 15:48
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    $\begingroup$ Using the profinite completion of the free abelian group as suggested by @YCor, seems a simpler solution. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 25, 2019 at 17:59
  • $\begingroup$ @ProfiniteQuestioner Where are you learning about profinite groups and profinite sets? Do they not cover such facts? My reference is Adamek and Rosicky's Locally Presentable and Accessible Categories, though that's not the most direct reference $\endgroup$
    – Tim Campion
    Commented Dec 26, 2019 at 5:16

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