# Is $\pi_1^{et}(\mathrm{Spec}\, k) \simeq \mathrm{Gal}(k^{sep}/k)$? (confusion about profinite completions)

The question is in the title, namely:

Suppose that $k$ is a field and $k^{sep}$ its absolute algebraic separable closure. Denote $\bar s :\mathrm{Spec}\, k^{sep} \rightarrow \mathrm{Spec}\, k$ the corresponding geometric point. Is $\pi_1^{et}(\mathrm{Spec}\, k, \bar s ) \simeq \mathrm{Gal}(k^{sep}/k)$ (as topological groups)?

This seems really as something that should hold, since this is a motivating example of étale fundamental groups. Let me describe where is my problem.

At the moment, I hold the following conflicting believes:

1. $\pi_1^{et}(\mathrm{Spec}\, k, \bar s) \simeq \mathrm{Gal}(k^{sep}/k)$, as stated.

2. $\pi_1^{et}(\mathrm{Spec}\, k, \bar s)$ is given as the group of automorphisms of the fibre functor $F_{\bar s}: \mathsf{FEt}_{\mathrm{Spec}\, k} \rightarrow \mathsf{Sets}, X \mapsto X_{\bar s}=X \times_k\mathrm{Spec}\,k^{sep}.$ Moreover (more importantly), $(\mathsf{FEt}_{\mathrm{Spec}\, k}, F_{\bar s})$ is a Galois category, e.g. as in Definition 52.3.6 on Stacksproject.

3. There is an equivalence of Galois categories $(\mathsf{FEt}_{\mathrm{Spec}\, k}, F_{\bar s})$ and $(G\mathsf{-sets}, U)$, where $G=Aut(F_{\bar s}),$ $G\mathsf{-sets}$ denotes the cat. of all finite $G$-sets with continuous $G$-action and $U:G\mathsf{-sets}\rightarrow \mathsf{Sets}$ is the forgetful functor (cf. e.g. Proposition 52.3.10 on Stacksproject). In particular, $Aut(F_{\bar s})=Aut(U)$ as topological groups.
4. Given a topological group $G$ and $U: G\mathsf{-sets} \rightarrow \mathsf{Sets}$ the forgetful functor, $Aut(U) \simeq \widehat G$, the profinite completion of $G$, as in Lemma 52.3.3 on Stacksproject.
5. The group $\mathrm{Gal}(k^{sep}/k)$ is not necessarily its own profinite completion, because there are in general subgroups of finite index that are not open. I believe this is shown e.g. in Milne's field theory notes (chapter 7, Proposition 7.26) in the case $k=\mathbb{Q}$.

It seems to me that using 2.-4., one obtains that $$\pi_1^{et}(\mathrm{Spec}\, k, \bar s ) \simeq \widehat{\mathrm{Gal}(k^{sep}/k)},$$ which by 5. is in general not isomorphic to $\mathrm{Gal}(k^{sep}/k).$ That is in conflict with 1.

So more specifically:

Which one of the above statements is wrong? If neither one is, why is the conclusion wrong?

Thanks in advance for any help.

Your error is in the conclusion. In statement (4), $\widehat{G}$ is the profinite completion of $G$ as a topological group. That means you take the completion with respect to only continuous homorphisms from $G$ to finite groups, not all homomorphisms. Any profinite group with its profinite topology is its own profinite completion as a topological group.