Lemma 58.3.11. Let $(\mathcal{C}, F)$ and $(\mathcal{C}', F')$ be Galois categories. Let $H : \mathcal{C} \to \mathcal{C}'$ be an exact functor. There exists an isomorphism $t : F' \circ H \to F$. The choice of $t$ determines a continuous homomorphism $h : G' = \text{Aut}(F') \to \text{Aut}(F) = G$ and a $2$-commutative diagram

$\xymatrix{ \mathcal{C} \ar[r]_ H \ar[d] & \mathcal{C}' \ar[d] \\ \textit{Finite-}G\textit{-Sets} \ar[r]^ h & \textit{Finite-}G'\textit{-Sets} }$

The map $h$ is independent of $t$ up to an inner automorphism of $G$. Conversely, given a continuous homomorphism $h : G' \to G$ there is an exact functor $H : \mathcal{C} \to \mathcal{C}'$ and an isomorphism $t$ recovering $h$ as above.

Proof. By Proposition 58.3.10 and Lemma 58.3.3 we may assume $\mathcal{C} = \textit{Finite-}G\textit{-Sets}$ and $F$ is the forgetful functor and similarly for $\mathcal{C}'$. Thus the existence of $t$ follows from Lemma 58.3.4. The map $h$ comes from transport of structure via $t$. The commutativity of the diagram is obvious. Uniqueness of $h$ up to inner conjugation by an element of $G$ comes from the fact that the choice of $t$ is unique up to an element of $G$. The final statement is straightforward. $\square$

Comment #2059 by Taylor Dupuy on

small typo: "innner" third to last line of the proof (html version).

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