Lemma 57.4.3. In diagram (57.4.0.1) the following are equivalent

1. the sequence $G'' \xrightarrow {h'} G' \xrightarrow {h} G \to 1$ is exact in the following sense: $h$ is surjective, $h \circ h'$ is trivial, and $\mathop{\mathrm{Ker}}(h)$ is the smallest closed normal subgroup containing $\mathop{\mathrm{Im}}(h')$,

2. $H$ is fully faithful and an object $X'$ of $\mathcal{C}'$ is in the essential image of $H$ if and only if $H'(X')$ is isomorphic to a finite coproduct of final objects, and

3. $H$ is fully faithful, $H \circ H'$ sends every object to a finite coproduct of final objects, and for an object $X'$ of $\mathcal{C}'$ such that $H'(X')$ is a finite coproduct of final objects there exists an object $X$ of $\mathcal{C}$ and an epimorphism $H(X) \to X'$.

Proof. By Lemmas 57.4.1 and 57.4.2 we may assume that $H$ is fully faithful, $h$ is surjective, $H' \circ H$ maps objects to disjoint unions of the final object, and $h \circ h'$ is trivial. Let $N \subset G'$ be the smallest closed normal subgroup containing the image of $h'$. It is clear that $N \subset \mathop{\mathrm{Ker}}(h)$. We may assume the functors $H$ and $H'$ are the canonical functors $\textit{Finite-}G\textit{-Sets} \to \textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$ determined by $h$ and $h'$.

Suppose that (2) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, the action of $G'$ factors through $G$. Apply this to $X' = G'/U'N$ where $U'$ is a small open subgroup of $G'$. Then we see that $\mathop{\mathrm{Ker}}(h) \subset U'N$ for all $U'$. Since $N$ is closed this implies $\mathop{\mathrm{Ker}}(h) \subset N$, i.e., (1) holds.

Suppose that (1) holds. This means that $N = \mathop{\mathrm{Ker}}(h)$. Let $X'$ be a finite $G'$-set such that $G''$ acts trivially. This means that $\mathop{\mathrm{Ker}}(G' \to \text{Aut}(X'))$ is a closed normal subgroup containing $\mathop{\mathrm{Im}}(h')$. Hence $N = \mathop{\mathrm{Ker}}(h)$ is contained in it and the $G'$-action on $X'$ factors through $G$, i.e., (2) holds.

Suppose that (3) holds. This means that for a finite $G'$-set $X'$ such that $G''$ acts trivially, there is a surjection of $G'$-sets $X \to X'$ where $X$ is a $G$-set. Clearly this means the action of $G'$ on $X'$ factors through $G$, i.e., (2) holds.

The implication (2) $\Rightarrow$ (3) is immediate. This finishes the proof. $\square$

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