# Etale Fundamental Groups of Spec k

I am studying etale fundamental groups, and I am confused with the basic example: $$\pi_1(\text{Spec}\ K)$$ for a field $$K$$. I tried to read several explanations of this example, but none of them provided a very detailed answer to this question.

Choose a geometric point $$\overline{x}: \text{Spec }\overline{K}\rightarrow \text{Spec }K$$, or equivalently an embedding $$K\hookrightarrow \overline{K}$$. The "Galois coverings" of the scheme $$\text{Spec }K$$ together with the geometric point $$\overline{x}$$ are just the morphisms $$\text{Spec }L\rightarrow \text{Spec }K$$, where $$L$$ is a finite Galois extension of $$K$$. The fiber of this covering over $$\overline{x}$$ is the set of geometric points $$\overline{y}: \text{Spec }\overline{K}\rightarrow \text{Spec }L$$ which lies over $$\overline{x}$$, or equivalently, the field embeddings $$L\hookrightarrow \overline{K}$$ which extends the embedding $$K\hookrightarrow \overline{K}$$ corresponding to $$\overline{x}$$. Once we fix an embedding $$K\hookrightarrow \overline{K}$$, the image in $$\overline{K}$$ of the finite Galois extension $$L$$ is also fixed regardless of the choice of the embedding. Hence, the fiber $$Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$$ is in one-to-one correspondence with the Galois group $$Gal(L/K)$$. The etale fundamental group $$\pi_1(\text{Spec }K, \overline{x})$$ is, by definition, the automorphism group of the fiber functor $$Fib_\overline{x}$$ from the category of (finite) etale coverings of $$\text{Spec }K$$ to the category of sets.

I know that, by definition, an automorphism of $$Fib_\overline{x}$$ is a collection of automorphisms (in the category of sets; so just bijections) of $$Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$$ which is compatible with the morphisms $$Fib_\overline{x}(\text{Spec }L_1 \rightarrow \text{Spec }L_2)$$ coming from each pair of finite Galois extensions $$(L_1, L_2)$$ of $$K$$ and a field homomorphism $$L_2\rightarrow L_1$$ between them. This morphism $$Fib_\overline{x}(\text{Spec }L_1 \rightarrow \text{Spec }L_2)$$ maps $$\overline{y}:\text{Spec }\overline{K}\rightarrow \text{Spec }L_1$$ to $$\overline{y}': \text{Spec }\overline{K}\rightarrow \text{Spec }L_1\rightarrow \text{Spec }L_2$$. Hence, an automorphism $$\alpha$$ maps each embedding $$\iota: L\hookrightarrow \overline{K}$$ to another embedding $$\alpha_L(\iota):L\hookrightarrow \overline{K}$$, and if $$\varphi:L'\rightarrow L$$ is an embedding of another finite Galois extension $$L'$$ into $$L$$, then $$\alpha_{L'}(\iota\circ \varphi) = \alpha_{L}(\iota)\circ \varphi$$.

Now I am stuck here: obviously the elements of the absolute Galois group $$Gal(K^{sep}/K)$$ give such automorphisms, but conversely how do I show that this automorphism $$\alpha$$ is actually coming from an element of $$Gal(K^{sep}/K)$$? In other words, why is the map $$\alpha_L:Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)\rightarrow Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$$ not just a bijection of sets (i.e. an element of the symmetric group $$S_{[L:K]}$$) but actually a composition with an element of $$Gal(L/K)$$?

Actually, while writing this question, I could think a possible strategy: focus on the actual subfields $$L$$ of $$\overline{K}$$ (viewing $$K$$ as a subfield of $$\overline{K}$$), and use the identification of $$Fib_\overline{x}(\text{Spec }L\rightarrow \text{Spec }K)$$ with $$Gal(L/K)$$ as above, to assign to each $$\alpha\in Aut(Fib_\overline{x})$$ an element of $$Gal(K^{sep}/K)$$. So we can identify the inclusion $$\iota_0:L\hookrightarrow \overline{K}$$ with the identity element of $$Gal(L/K)$$. Then $$\alpha_L(\iota_0)$$ is another element of $$Gal(L/K)$$. If $$\varphi_0: L'\hookrightarrow L$$ is the inclusion, then $$\alpha_{L'}(\iota_0\circ\varphi_0) = \alpha_L(\iota_0)\circ\varphi_0$$ is the restriction of $$\alpha_L(\iota_0)\in Gal(L/K)$$ to $$L'$$ (as $$L'$$ is Galois over $$K$$ so $$\alpha_L(\iota_0)(L')=L'$$). So $$\alpha$$ at each inclusion of the subfields of $$\overline{K}$$ which are finite Galois over $$K$$ defines a collection of elements of the Galois groups $$Gal(L/K)$$ for such extensions $$L$$, and this elements are compatible with restrictions. Therefore they can be glued to define an element of $$Aut(\overline{K}/K) =Gal(K^{sep}/K)$$.

Does this argument actually work? Also, is everything I wrote correct?