Lemma 58.4.4. In diagram (58.4.0.1) the following are equivalent

1. $h'$ is injective, and

2. for every connected object $X''$ of $\mathcal{C}''$ there exists an object $X'$ of $\mathcal{C}'$ and a diagram

$X'' \leftarrow Y'' \rightarrow H(X')$

in $\mathcal{C}''$ where $Y'' \to X''$ is an epimorphism and $Y'' \to H(X')$ is a monomorphism.

Proof. We may replace $H'$ by the corresponding functor between the categories of finite $G'$-sets and finite $G''$-sets.

Assume $h' : G'' \to G'$ is injective. Let $H'' \subset G''$ be an open subgroup. Since the topology on $G''$ is the induced topology from $G'$ there exists an open subgroup $H' \subset G'$ such that $(h')^{-1}(H') \subset H''$. Then the desired diagram is

$G''/H'' \leftarrow G''/(h')^{-1}(H') \rightarrow G'/H'$

Conversely, assume (2) holds for the functor $\textit{Finite-}G'\textit{-Sets} \to \textit{Finite-}G''\textit{-Sets}$. Let $g'' \in \mathop{\mathrm{Ker}}(h')$. Pick any open subgroup $H'' \subset G''$. By assumption there exists a finite $G'$-set $X'$ and a diagram

$G''/H'' \leftarrow Y'' \rightarrow X'$

of $G''$-sets with the left arrow surjective and the right arrow injective. Since $g''$ is in the kernel of $h'$ we see that $g''$ acts trivially on $X'$. Hence $g''$ acts trivially on $Y''$ and hence trivially on $G''/H''$. Thus $g'' \in H''$. As this holds for all open subgroups we conclude that $g''$ is the identity element as desired. $\square$

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