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.



Is there a free profinite abelian group on a profinite set?
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](https://ncatlab.org/nlab/show/pro-object) $Pro(Ab^{fin})$ and $Pro(Set^{fin})$ are complete and cocomplete. In fact, they are opposite to [locally finitely presentable categories](https://ncatlab.org/nlab/show/locally+finitely+presentable+category) -- 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](https://ncatlab.org/nlab/show/free+cocompletion) $U^\ast: Fun(Set^{fin},Set) \to Fun(Ab^{fin}, Set)$ 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.


