## 58.6 Fundamental groups

In this section we define Grothendieck's algebraic fundamental group. The following definition makes sense thanks to Lemma 58.5.5.

Definition 58.6.1. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$. The *fundamental group* of $X$ with *base point* $\overline{x}$ is the group

\[ \pi _1(X, \overline{x}) = \text{Aut}(F_{\overline{x}}) \]

of automorphisms of the fibre functor $F_{\overline{x}} : \textit{FÉt}_ X \to \textit{Sets}$ endowed with its canonical profinite topology from Lemma 58.3.1.

Combining the above with the material from Section 58.3 we obtain the following theorem.

Theorem 58.6.2. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$.

The fibre functor $F_{\overline{x}}$ defines an equivalence of categories

\[ \textit{FÉt}_ X \longrightarrow \textit{Finite-}\pi _1(X, \overline{x})\textit{-Sets} \]

Given a second geometric point $\overline{x}'$ of $X$ there exists an isomorphism $t : F_{\overline{x}} \to F_{\overline{x}'}$. This gives an isomorphism $\pi _1(X, \overline{x}) \to \pi _1(X, \overline{x}')$ compatible with the equivalences in (1). This isomorphism is independent of $t$ up to inner conjugation.

Given a morphism $f : X \to Y$ of connected schemes denote $\overline{y} = f \circ \overline{x}$. There is a canonical continuous homomorphism

\[ f_* : \pi _1(X, \overline{x}) \to \pi _1(Y, \overline{y}) \]

such that the diagram

\[ \xymatrix{ \textit{FÉt}_ Y \ar[r]_{\text{base change}} \ar[d]_{F_{\overline{y}}} & \textit{FÉt}_ X \ar[d]^{F_{\overline{x}}} \\ \textit{Finite-}\pi _1(Y, \overline{y})\textit{-Sets} \ar[r]^{f_*} & \textit{Finite-}\pi _1(X, \overline{x})\textit{-Sets} } \]

is commutative.

**Proof.**
Part (1) follows from Lemma 58.5.5 and Proposition 58.3.10. Part (2) is a special case of Lemma 58.3.11. For part (3) observe that the diagram

\[ \xymatrix{ \textit{FÉt}_ Y \ar[r] \ar[d]_{F_{\overline{y}}} & \textit{FÉt}_ X \ar[d]^{F_{\overline{x}}} \\ \textit{Sets} \ar@{=}[r] & \textit{Sets} } \]

is commutative (actually commutative, not just $2$-commutative) because $\overline{y} = f \circ \overline{x}$. Hence we can apply Lemma 58.3.11 with the implied transformation of functors to get (3).
$\square$

Lemma 58.6.3. Let $K$ be a field and set $X = \mathop{\mathrm{Spec}}(K)$. Let $\overline{K}$ be an algebraic closure and denote $\overline{x} : \mathop{\mathrm{Spec}}(\overline{K}) \to X$ the corresponding geometric point. Let $K^{sep} \subset \overline{K}$ be the separable algebraic closure.

The functor of Lemma 58.2.2 induces an equivalence

\[ \textit{FÉt}_ X \longrightarrow \textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets}. \]

compatible with $F_{\overline{x}}$ and the functor $\textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets} \to \textit{Sets}$.

This induces a canonical isomorphism

\[ \text{Gal}(K^{sep}/K) \longrightarrow \pi _1(X, \overline{x}) \]

of profinite topological groups.

**Proof.**
The functor of Lemma 58.2.2 is the same as the functor $F_{\overline{x}}$ because for any $Y$ étale over $X$ we have

\[ \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(\overline{K}), Y) = \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(K^{sep}), Y) \]

Namely, as seen in the proof of Lemma 58.2.2 we have $Y = \coprod _{i \in I} \mathop{\mathrm{Spec}}(L_ i)$ with $L_ i/K$ finite separable over $K$. Hence any $K$-algebra homomorphism $L_ i \to \overline{K}$ factors through $K^{sep}$. Also, note that $F_{\overline{x}}(Y)$ is finite if and only if $I$ is finite if and only if $Y \to X$ is finite étale. This proves (1).

Part (2) is a formal consequence of (1), Lemma 58.3.11, and Lemma 58.3.3. (Please also see the remark below.)
$\square$

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