MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
Are you familiar with the adjoint functor theorem? – S. Carnahan Jun 26 '13 at 20:42
up vote 5 down vote accepted

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.

share|cite|improve this answer
In fact, the corresponding functor is accessible, so it follows directly from the accessible adjoint functor theorem. – Zhen Lin Jun 27 '13 at 3:12
Thanks, I learnt a lot from this answer. – Jason Polak Jun 27 '13 at 18:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.