Let $G$ be a group and let $\mathbb{G}$ be the associated one object category. Is there an explicit presentation of representable functors from $\mathbb{G} \to $Set? If so how does the Yoneda lemma look like explicitly in this setting?
-
2$\begingroup$ I think this question is addressed here: math.stackexchange.com/questions/2456429/… $\endgroup$– Ben MacAdamCommented Mar 1, 2023 at 17:18
-
1$\begingroup$ It’s a free G-set. Yoneda lemma says that a map to a G-set from a free one is just an ordinary map from the basis. $\endgroup$– Fernando MuroCommented Mar 1, 2023 at 19:23
-
1$\begingroup$ Approximately like Cayley's theorem that every group is a subgroup of the permutation group of its underlying set. :-) $\endgroup$– David Roberts ♦Commented Mar 2, 2023 at 6:07
1 Answer
Let me write $\def\B{\mathbf{B}}\B G$ for what you call $\mathbb{G}$.
You can check that presheaves on $\B G$ are precisely the right $G$-sets: the unique point of $\B G$ is sent to some set $X$, and functoriality defines a group homomorphism $\def\op{\mathrm{op}}G^\op\to\operatorname{Aut}(X)$.
In particular, the (unique) representable functor corresponds to the $G$-set $G$ itself with the action given by right multiplication.
The Yoneda Lemma in this setting then says, for any $G$-set $X$, that elements of $X$ correspond naturally to $G$-equivariant maps $G\to X$: send $x\in X$ to the map $g\mapsto x.g$, and send a function $f:G\to X$ to $f(1_G)$.
Edit: to be a bit more explicit about the naturality, the group $G$ acts on both sides of this correspondence (the action on $X$ is given by the fact that $X$ is a $G$-set, and the action on $G$-equivariant functions $G\to X$ is componentwise), and the naturality of the Yoneda lemma says that this correspondence respects these $G$-actions.
-
$\begingroup$ What does "naturally" mean explicitly in this setting? $\endgroup$ Commented Mar 1, 2023 at 19:20
-
$\begingroup$ @QuinAppleby "correspond naturally" means that the proof of the Yoneda lemma gives a bijection between between both sides $\endgroup$ Commented Mar 1, 2023 at 20:37
-
$\begingroup$ @QuinAppleby I have made an edit to elaborate on the naturality in this setting. $\endgroup$ Commented Mar 1, 2023 at 20:52