Clarification in Lenstra's Galois Theory for Schemes notes

Question

I have a question about subsection 3.17 in Lenstra's Galois Theory for Schemes notes.

It says that $\operatorname{Aut}_\mathbf{C}(A)$ acts transitively on $\operatorname{Mor}_{\mathbf{C}}(A,B)$ (via pre-composition). I get why this implies that $$H(B)=F(B)\cong \operatorname{Mor}_{\mathbf{C}}(A,B) \cong \operatorname{Aut}_{\mathbf{C}}(A)/G$$ (where $G$ is the stabilizer of any $f:A\to B$) as $\operatorname{Aut}_{\mathbf{C}}(A)$-sets but why does this mean they are also isomorphic as $\pi$-sets?

Some context/a refresher: $\mathbf{C}$ is a Galois category, $A\in \mathbf{C}$, $B\in \mathbf{C}$ is a connected object (as defined in 3.12), $F:\mathbf{C}\to \operatorname{FiniteSets}$ is a functor (with some properties, see 3.1), $\pi$ is the profinite group defined in 3.15.

Thanks!

Answer 1

I think I got this and it's essentially a tautology but I'll write it up explicitly. Since $H(B)=F(B)=\operatorname{Mor}_{\mathbf{C}}(A,B)$ as sets, I will just refer to the latter. Now there is an action of $\operatorname{Aut}_{\mathbf{C}}(A)$ on $\operatorname{Mor}_\mathbf{C}(A,B)$ given by $\sigma \cdot f := f\circ \sigma^{-1}$. In Lenstra's notes, it's shown that this is a transitive action thus we have an isomorphism $\operatorname{Mor}_\mathbf{C}(A,B) \cong \operatorname{Aut}_{\mathbf{C}}(A)/G$ where $G=\{\sigma \mid \sigma \cdot f = f\}$ amd the isomorphism is explicitly given by $\sigma \cdot f \mapsto \sigma G$ where $f\in \operatorname{Mor}_\mathbf{C}(A,B)$ is fixed. This is an isomorphism of $\operatorname{Aut}_{\mathbf{C}}(A)$-sets but I claim it's also $\pi$-equivariant so it will be an isomorphism of $\pi$-sets.

Now let's recall what $\pi$ is and how it acts on the two objects above. By definition we have $\pi = \varprojlim_{J} \operatorname{Aut}_{\mathbf{C}}(D)$ so an element in $\pi$ is a collection $(\sigma_D)_{D\in J}$ (I'm supressing the notation $(D,d)\in J$) that are compatible under the restriction maps of the inverse limit. Checking the definition, this acts on $\operatorname{Mor}_\mathbf{C}(A,B)$ by $(\sigma_D)_D\cdot f:= \sigma_A\cdot f = f\circ \sigma_A^{-1}$. In other words, it's exactly the action of $\operatorname{Aut}_{\mathbf{C}}(A)$. We also get an action of $\pi$ on $\operatorname{Aut}_{\mathbf{C}}(A)/G$ by $(\sigma_D)_{D}\cdot (\sigma G):=(\sigma_A \circ \sigma) G$.

Now we check the isomorphism (call it $\psi$) in the first paragraph is $\pi$-equivariant. We have to check that $$\psi((\sigma_D)_D \cdot (\sigma\cdot f))=(\sigma_D)_D \cdot \psi(\sigma \cdot f)$$ i.e. $\psi(\sigma_A\cdot (\sigma\cdot f)=\sigma_A \cdot \psi(\sigma \cdot f)$ which is true as $\psi$ is $\operatorname{Aut}_{\mathbf{C}}(A)$-equivariant so here's why this was really just a tautology.