Let $G$ be a group, and $\mathbf{G\text{-}Sets}$ the category whose objects are left G-Sets and whose morphisms are G-Set homomorphisms, that is functions $f:X\to Y$ such that $f(ax) = af(x)$, $a\in G, x\in X$.

Does the forgetful functor $\mathbf{G\text{-}Sets}$ $\to$ $\mathbf{Sets}$ have a left-adjoint?

Are there any other interesting functors between the categories $\mathbf{G\text{-}Sets}$ and $\mathbf{Sets}$?


Let $S$ be a set. Then $G \times S$ can be equipped with a $G$-action by having $G$ act by left-multiplication on itself and trivially on $S$. This is a left-adjoint to the forgetful functor -- for $T$ an arbitrary $G$-set, one has

$$ \text{Hom}_\text{G-set}(G \times S, T) = \text{Hom}_\text{set}(S, T) $$ by the rule $f \mapsto \left[s \mapsto f(1, s)\right]$. The map the other way is $\phi \mapsto \left[(g, s) \mapsto g\phi(s)\right]$ .

  • $\begingroup$ Why did you write $\text{Hom}_\text{G-set}(G \times S, T)$? Shouldn't it be $\text{Hom}_\text{G-set}(S, T)$ where $S$ is a set equipped with a certain map $G\times S \to S$? Those are the objects in G-Set -Thanks $\endgroup$ – iwriteonbananas Apr 21 '15 at 17:18
  • $\begingroup$ my functor eats $S$ and returns $G \times S$ with the action described in the answer. That's what it means to be a left adjoint. In other words, if I call the functor $F$, the statement is $Hom_{G-set}(F(S), T) = Hom_{set}(S, Forget(T))$. $\endgroup$ – hunter Apr 21 '15 at 21:52

This is an instance of Kan extensions.

Let $\mathbf G$ be the one object category associated to the group $G$, and $\mathbf 1$ the one associated to the trivial group. Then $\mathbf{G{-}Sets}$ is nothing else than $[\mathbf G,\mathbf{Sets}]$ and the forgetful functor is the functor $$ i^\ast \colon [\mathbf G,\mathbf{Sets}] \to [\mathbf 1,\mathbf{Sets}] \simeq \mathbf{Sets},\quad F \mapsto F \circ i $$ induced by the (unique) functor $i\colon \mathbf 1 \to \mathbf G$. As $\mathbf{Sets}$ is cocomplete, the functor $i^\ast$ admits a left adjoint, usually denoted $i_!$ (see nlab:Kan extension). Note that, as $\mathbf{Sets}$ is also complete, $i^\ast$ also admits a right adjoint, usually denoted $i_\ast$.

Remark that by computing the general formula for left Kan extension in this case, you get back hunter's answer: $$ i_! \colon S \mapsto \coprod_{g\in G}S \simeq G \times S. $$


Here's another way of seeing why this functor has a left-adjoint.

Theorem. Let $\mathsf{S}$ and $\mathsf{T}$ denote Lawvere theories. Then given a morphism $\varphi : \mathsf{S} \leftarrow \mathsf{T}$, the corresponding forgetful functor $$\mathrm{Mod}(\varphi) : \mathrm{Mod}(\mathsf{T}) \leftarrow \mathrm{Mod}(\mathsf{S})$$ has a left-adjoint.


  • The initial Lawvere theory is denoted $\mathsf{Set}$, and its models are (basically) sets.
  • Given a group $G,$ we can build a Lawvere theory $\mathrm{Lath}(G)$ such that the models of $\mathrm{Lath}(G)$ are (basically) precisely the $G$-sets.
  • Write $\varphi$ for the unique morphism $\varphi : \mathrm{Lath}(G) \leftarrow \mathsf{Set}.$

So use the above theorem to conclude that

$$\mathrm{Mod}(\varphi) : \mathrm{Mod}(\mathsf{Set}) \leftarrow \mathrm{Mod}(\mathrm{Lath}(G))$$

has a left-adjoint. In other words, the usual forgetful functor

$$\mathbf{Set} \leftarrow \mathbf{Set}^G$$

has a left-adjoint, where I write $\mathbf{Set}^G$ for the category of $G$-sets, since a $G$-set is the same thing as a functor $\mathbf{Set} \leftarrow G,$ where $G$ is viewed as a category with one object.

Although this uses a lot more technical machinery than is strictly required, I like this viewpoint because it provides a quick heuristic that can help you immediately conclude that a variety of different functors have left-adjoints.


If $H\le G$ is a subgroup let $\mathrm{Res}_H^G:G\textrm{-}\mathsf{Set}\to H\textrm{-}\mathsf{Set}$ be the restriction functor. One can think of $\mathsf{Set}$ as the case where $H=1$ is the trivial subgroup. It has a left adjoint, call it $\mathrm{Ind}_H^G$, so that

$$\hom_{G\textrm{-}\mathsf{Set}}(\mathrm{Ind}_H^G-,-) ~~\cong~ \hom_{H\textrm{-}\mathsf{Set}}(-,\mathrm{Res}_H^G-)$$

are naturally isomorphic functors $H\textrm{-}\mathsf{Set}\times G\textrm{-}\mathsf{Set}\to\mathsf{Set}$.

An explicit construction is $\mathrm{Ind}_H^G\Omega\cong G\times_H \Omega$, which is $G\times\Omega$ (the action of $G$ purely on the first coordinate) modulo the relation $(gh,\omega)\sim(g,h\omega)$ for all $g\in G,h\in H,\omega\in\Omega$.

Does this look familiar? If you've studied representation theory, it should. The linearized version of this fact is called Frobeius reciprocity. Let $G\textrm{-}\mathsf{Rep}$ be the category of linear representations of $G$, and then define the restriction functor $\mathrm{Res}_H^G:G\textrm{-}\mathsf{Rep}\to H\textrm{-}\mathsf{Rep}$. Frobenius reciprocity (the categorical version of it, as opposed to the more standard, decategorified, character-theoretic version of it) states that restriction has a left adjoint called induction $\mathrm{Ind}_H^G$, so $\hom_G(\mathrm{Ind}_H^GV,W)\cong\hom_H(V,\mathrm{Res}_H^GW)$ canonically for all representations $V$ of $G$ and $W$ of $H$.

An explicit construction is $\mathrm{Ind}_H^GV\cong \Bbb C[G]\otimes_{\Bbb C[H]}V$, the linearized analogue of $G\times_H\Omega$!


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