A group action seen as a functor is representable iff for every $x,y \in A$ there is exactly one $g \in G$ such that $gx=y$.

Question

If you have a group action form $G$ to $A$ seen as functor $F:G \rightarrow Set$ then it has a universal element if and only if for every $x,y \in A$ there is exactly one element in $g \in G$ such that $gx=y$.

This is a consequence of Yoneda's lemma but I can't see why. Any help with this will be greatly appreciated.

Answer 1

The functor $F : G \to \mathbf{Set}$ is representable if and only if there is a natural isomorphism $$\theta : F \overset{\cong}{\longrightarrow} G(\star, {-})$$ where $\star$ is the unique object of $G$.

Let's investigate what this means.

• Since $G$ only has one object $\star$, the natural transformation $\theta$ is determined by its unique component $\theta_{\star} : F(\star) \to G(\star,\star)$, which is a function $A \to G$;
• The fact that $\theta$ is invertible says precisely that the map $\theta_{\star}$ is a bijection, and so the set $A$ is in bijection with (the underlying set of) the group $G$;
• Naturality of $\theta$ says that for every $g \in G$, we have $g\theta_{\star}(a) = \theta_{\star}(g \cdot a)$ for all $a \in A$.

So let $x,y \in A$ and let $\theta_{\star}(x)=g \in G$ and $\theta_{\star}(y)=h \in G$. Then $$\theta_{\star}(hg^{-1} \cdot x) = hg^{-1} \theta_{\star}(x) = hg^{-1}g = h = \theta_{\star}(y)$$ The fact that $\theta_{\star}$ is invertible gives you that $hg^{-1} \cdot x = y$.