Deducing Cayley's theorem from the Yoneda lemma using Wikipedia's recipe

Question

I'm following Wikipedia in trying to prove that Cayley's theorem emerges as a particular case of the Yoneda lemma. In case that article gets edited, here's the screenshot:

A couple of aspects in the proof are unclear:

• In proving that the set of $G$-equivariant maps $\alpha_\ast:\mathcal C(\ast,\ast)\to\mathcal C(\ast,\ast)$ is a group, I'm not sure what the inverse of $\alpha_\ast$ is.

• A (probably) related question: $X$ is the image of the unique object $\ast$ under $H^\ast$, i.e., $\mathcal C(\ast,\ast)$. $\text{Perm}(X)$ is the set/group of bijections from $\mathcal C(\ast,\ast)$ into itself. To show that all equivariant maps form a subgroup of this group, we need to show that each equivariant map $\alpha_\ast$ is a bijection from $\mathcal C(\ast,\ast)$ into itself. But I don't see why $\alpha_\ast$ has to be a bijection. $\alpha$ is just a natural transformation, not a natural isomorphism.

• I'm a little confused about the this point from Wikipedia: "(2) the function which gives the bijection is a group homomorphism". What function is meant here? The function from the (dual version of the) Yoneda lemma $[\mathcal C,\textbf{Set}](H^\ast,H^\ast)\to\mathcal C(\ast,\ast)$ doesn't seem to be exactly the right function because its domain has elements that are natural transformations $\alpha$, but we want to construct a function from the set of $G$-equivariant maps, which have the form $\alpha_\ast$ for some natural transformation $\alpha$. So I suspect that the map that is meant is the composite $$\{\alpha_\ast: \alpha:H^\ast\to H^\ast\text{ is a nat. transf.}\}\to \{\text{nat. transf. } \alpha:H^\ast\to H^\ast\}\to \mathcal C(\ast,\ast)$$ where the first map is a "natural" bijection (natural in the non-technical sense). But if so, I don't really see why this is a group homomorphism. If we call this composition $f$, then we need to show that $f(\beta_\ast\circ\alpha_\ast)=f(\beta_\ast)\circ f(\alpha_\ast)$ or equivalently $\beta_\ast(\alpha_\ast(1_\ast))=\beta_\ast(1_\ast)\circ\alpha_\ast(1_\ast)$. I don't really see why this is true, although this must be something easy.

Answer 1

• a $G$-equivariant map $\alpha : G \to G$ for this action satisfies $\alpha(g) = g\alpha(e)$, where $e$ is the identity in $G$, so $\alpha$ is just right-multiplication by $\alpha(e)$, so it is a bijection and its inverse is right-multiplication by $\alpha(e)^{-1}$.
• Assuming $H^*$ means $\mathcal{C}(*, -)$, the reason the $G$-equivariant maps on $\mathcal{C}(*, *)=G$ are bijections is that they turn out to be right-multiplication by a group element.
• the article is presumably referring to the isomorphism $\hom(\hom_\mathcal{C}(*, -), F) \to F(*)$ provided by Yoneda after making the identification they describe in the previous paragraph of the left hand side with the set of equivariant maps $G \to G$. A natural transformation $\beta$ regarded as an equivariant map gets sent to $\beta(e)$, so the composition $\alpha \circ \beta$ goes to $\alpha(\beta(e)) = \alpha(\beta(e) e) = \beta(e)\alpha(e)$. Huh.

The latter looks wrong because, unless I have things mixed up, the wiki article isn't quite correct. It says that something is a group homomorphism and then adds that "going in the reverse direction, it associates to every $g$ in $G$ the equivariant map of right-multiplication by $g$". But that is an antihomomophism not a homomorphism: $gh$ goes to the right multiplication map $r_{gh}(x) = xgh$ which is $r_h\circ r_g$, not $r_g \circ r_h$.