Orbit functor is not co-representable

Question

Let ${1}\neq H\le G$ be groups. Denote by $G\textit{-}\mathsf{Set}$ the category of sets with a $G$ action, with $G$-equivariant maps as morphisms. Let $(-)/H: G\textit{-}\mathsf{Set}\to \mathsf{Set}$ be the functor sending every $G$-set $X$ to its set of $H$-orbits, and every $G$-equivariant map to its restriction to the set of orbits. Show that $(-)/H$ is not corepresentable.

It seems that this functor does preserve coproducts. A coproduct object of $X$ and $Y$ in $G\textit{-}\mathsf{Set}$ is the well-defined ("already existing") $G$ action on the disjoint union, $X\sqcup Y$. However, it is true (as far as I am concerned) that the orbits $(X\sqcup Y)/H$ in this action are precisely the disjoint union of orbits $X/H\sqcup Y/H $. So, this path did not yield a contadiction. I also tried to use explicitly a natural isomorphism to deduce that if $X_0$ is a corepresenting object, then $|X/H| = |\operatorname{Hom}_G (X_0, X)|$ for all $G$-sets $X$. Are these assesments correct? How to continue?

Answer 1

Let $|-|$ be the forgetful functor from $G$-sets to sets. There is a natural surjection $|-|\Rightarrow-/H$ whose components $|X|\twoheadrightarrow X/H$ send each element of $|X|$ to its $H$-orbit.

Having surjections $|X|\twoheadrightarrow\DeclareMathOperator\Hom{Hom}\Hom_G(X_0,X)$ would require that $\Hom_G(X_0,X)$ be non-empty whenever $|X|$ is non-empty. In that case, taking $X$ to be $G$ with its left action shows that $X_0\cong G\times|X_0/G|$. Then $\operatorname{Hom}_G(X_0,G)\cong |G|^{X_0/G}$ and naturality implies $|G|\twoheadrightarrow|G|^{X_0/G}$ is a surjective homomorphism for right $G$-actions, which requires $X_0/G$ be a singleton or empty.

In the case where $X_0/G$ is empty, $X_0$ is also empty, but then no $X_0\to\Hom_G(X_0,X_0)$ is surjective as the latter contains the identity homomorphism.

Thus we must have $X_0\cong G$, and then natural functions $|G|\to\Hom_G(G,G)=|G|$ are homomorphisms $G\to G$ for right $G$-actions, hence in bijective correspondence with points $e\in|G|$. Conversely, each $e\in|G|$ gives a natural bijection $|X|\twoheadrightarrow \Hom_G(G,X)$ sending $x\in|X|$ to the unique morphism $G\to X$ satisfying $e\mapsto x$. In other words, choices of points $e\in|G|$ are the data of $|-|$ being corepresented by $G$.

It follows that $-/H$ is corepresentable only if $|G|\cong G/H$, i.e. only if $H$ is the singleton subgroup. Conversely, if $H$ is the singleton group, then $-/H$ is isomorphic to the forgetful functor $|-|$, and hence corepresented by $G$ after a choice of $e$.