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Motivation

For a fixed group $G$, consider the category of $G$-groups, whose objects consist of (not necessarily abelian) groups $A$ together with an action $G\to \mathrm{Aut}(A)$. Now we can construct a nonabelian $H^1$, a functor from $G$-groups to pointed sets by mimicking the abelian construction. This construction is very common and there are Čech cohomology variants for instance. (Other variants of nonabelian cohomology $H^n(G,*)$ for all $n$ have also been constructed via crossed modules; see Labesse et al. in the Asterisque "Cohomologie, stabilisation et changement de base".)

One approach to try and construct higher $H^n(G,-)$ functors for the category of $G$-groups, would be to find a similar "nonabelian bar resolution". In the abelian case, the forgetful functor $U:G\text{-Mod}\to\text{Ab}$ has a left-adjoint $F:\text{Ab}\to G\text{-Mod}$ given by $A\mapsto \mathbb{Z}G\otimes_{\mathbb{Z}} A$. In this case we can then use the corresponding cotriple to construct the bar resolution in the usual fashion.

The Question:

A naive thing to do would be to look for a left adjoint to the forgetful functor $U:G\text{-Group}\to\text{Group}$, sending a $G$-group $A$ to the group $A$, forgetting the $G$-action. So my question is, does this forgetful functor have a left-adjoint? If so, what is it? My guess would be no, but I'm not sure. I've tried various things with free products but I can neither construct one nor prove it doesn't exist.

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    $\begingroup$ Are you familiar with the adjoint functor theorem? $\endgroup$
    – S. Carnahan
    Jun 26, 2013 at 20:42

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Yes. I believe there's a general theorem to the effect that if $T_1 \to T_2$ is any morphism of Lawvere theories, then the corresponding functor $\text{Prod}(T_2, \text{Set}) \to \text{Prod}(T_1, \text{Set})$ between the categories of models has a left adjoint (I think this follows from the general adjoint functor theorem).

$\text{Grp}$ is the category of models of a Lawvere theory given by taking all operations you can write down from the group axioms; more explicitly, it's the opposite of the category of free groups $F_n$. $G\text{-Grp}$ is the category of models of a more complicated Lawvere theory where, in addition to the operations you can write down from the group axioms, there is one additional unary operation for each element of $G$, and these operations satisfy all of the relations in $G$ and satisfy the relations making them group homomorphisms.

The general construction of the left adjoint is that one freely applies all possible operations in the Lawvere theory of $G\text{-Grp}$ to a group $A \in \text{Grp}$, then quotients by all relations coming from the fact that $A$ is already a group. Somewhat more explicitly, take the free monoid on symbols $(g_i a_i)$ where $g_i \in G, a_i \in A$, and quotient by all relations of the form $(g a)(g b) = (g (ab))$ and $(g \text{ id}_A) = (\text{id}_G \text{ id}_A)$. Multiplication is concatenation, inverses are given symbolwise by $(g a)^{-1} = (g a^{-1})$, the identity is $\text{id}_G \text{ id}_A$, and the $G$-action is given symbolwise by $g(ha) = (gh(a))$.

Edit: Equivalently, take the free product $\bigsqcup_{g \in G} gA$ over copies of $A$ indexed by $G$ (note that $\mathbb{Z}[G] \otimes A$ can be described analogously when $A$ is abelian) with $G$ acting on the indexing set.

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    $\begingroup$ In fact, the corresponding functor is accessible, so it follows directly from the accessible adjoint functor theorem. $\endgroup$
    – Zhen Lin
    Jun 27, 2013 at 3:12

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