The Stacks project

Lemma 58.4.1. In diagram ( the following are equivalent

  1. $h : G' \to G$ is surjective,

  2. $H : \mathcal{C} \to \mathcal{C}'$ is fully faithful,

  3. if $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is connected, then $H(X)$ is connected,

  4. if $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is connected and there is a morphism $*' \to H(X)$ in $\mathcal{C}'$, then there is a morphism $* \to X$, and

  5. for any object $X$ of $\mathcal{C}$ the map $\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(*, X) \to \mathop{\mathrm{Mor}}\nolimits _{\mathcal{C}'}(*', H(X))$ is bijective.

Here $*$ and $*'$ are final objects of $\mathcal{C}$ and $\mathcal{C}'$.

Proof. The implications (5) $\Rightarrow $ (4) and (2) $\Rightarrow $ (5) are clear.

Assume (3). Let $X$ be a connected object of $\mathcal{C}$ and let $*' \to H(X)$ be a morphism. Since $H(X)$ is connected by (3) we see that $*' \to H(X)$ is an isomorphism. Hence the $G'$-set corresponding to $H(X)$ has exactly one element, which means the $G$-set corresponding to $X$ has one element which means $X$ is isomorphic to the final object of $\mathcal{C}$, in particular there is a map $* \to X$. In this way we see that (3) $\Rightarrow $ (4).

If (1) is true, then the functor $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets}$ is fully faithful: in this case a map of $G$-sets commutes with the action of $G$ if and only if it commutes with the action of $G'$. Thus (1) $\Rightarrow $ (2).

If (1) is true, then for a $G$-set $X$ the $G$-orbits and $G'$-orbits agree. Thus (1) $\Rightarrow $ (3).

To finish the proof it suffices to show that (4) implies (1). If (1) is false, i.e., if $h$ is not surjective, then there is an open subgroup $U \subset G$ containing $h(G')$ which is not equal to $G$. Then the finite $G$-set $M = G/U$ has a transitive action but $G'$ has a fixed point. The object $X$ of $\mathcal{C}$ corresponding to $M$ would contradict (3). In this way we see that (3) $\Rightarrow $ (1) and the proof is complete. $\square$

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