Adjoints for the restriction of category-valued representations of groups

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Setup. Let $G$ be a group and let $\mathscr{A}$ be a category. We denote the category of functors from $G$ to $\mathscr{A}$ by $[G, \mathscr{A}]$ and think of these functors as $\mathscr{A}$-valued representation of $G$. (More explicitely, such a representation consists of an object $X$ of $\mathscr{A}$ and a homomorphism of groups from $G$ to $\operatorname{Aut}_{\mathscr{A}}(X)$.)

Let now $H$ be a subgroup of $G$. The inclusion map $i \colon H \to G$ can be regarded as a functor, which then induces a functor $$ \operatorname{res}^G_H ≔ i^* \colon [G, \mathscr{A}] \longrightarrow [H, \mathscr{A}] \,. $$ This functor restricts the $\mathscr{A}$-valued representations of $G$ to $\mathscr{A}$-valued representations of $H$.

Question. Under what conditions (on $\mathscr{A}$) does the restriction functor $\operatorname{res}^G_H$ admit a left adjoint, resp. a right adjoint?


This question is motivated by Exercise 2.1.16 in Tom Leinster’s Basic Category Theory, which deals with $[G, \mathrm{Set}]$ and $[G, \operatorname{Vect}(\mathbb{k})]$.

I believe that I understand the following two special cases:

  • For every ring $R$ we have for $\mathscr{A} = \operatorname{Mod}(R)$ that $[G, \mathscr{A}] ≅ \operatorname{Mod}(R[G])$. We thus have the usual adjunctions $$ R[G] \otimes_{R[H]} (-) ⊣ \operatorname{res}^G_H ⊣ \operatorname{Hom}_{R[H]}(R[G], -) \,. $$
  • In the case of $\mathscr{A} = \mathrm{Set}$ we have similarly $[G, \mathscr{A}] ≅ G\textrm{-}\mathrm{Set}$ and adjunctions $$ G \times_H (-) ⊣ \operatorname{res}^G_H ⊣ \operatorname{Hom}_H(G, -) \,. $$

However, I don’t expect these examples to generalize to an arbitrary category $\mathscr{A}$, as they both rely on some notion of tensor-hom adjuction.

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This is explained in Emily Riehl’s book “Category Theory in Context”, in particular in Example 6.2.8, as mentioned in the comments. More explicitely, a left adjoint exists if $\mathscr{A}$ is cocomplete, and a right adjoint exists if $\mathscr{A}$ is complete.