The Stacks project

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 $X$.

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

    \[ \textit{FÉt}_ X \longrightarrow \textit{Finite-}\pi _1(X, \overline{x})\textit{-Sets} \]
  2. 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.

  3. 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.

  1. 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 $\textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets} \to \textit{Sets}$.

  2. 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.) $\square$

Remark 58.6.4. In the situation of Lemma 58.6.3 let us give a more explicit construction of the isomorphism $\text{Gal}(K^{sep}/K) \to \pi _1(X, \overline{x}) = \text{Aut}(F_{\overline{x}})$. Observe that $\text{Gal}(K^{sep}/K) = \text{Aut}(\overline{K}/K)$ as $\overline{K}$ is the perfection of $K^{sep}$. Since $F_{\overline{x}}(Y) = \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(\overline{K}), Y)$ we may consider the map

\[ \text{Aut}(\overline{K}/K) \times F_{\overline{x}}(Y) \to F_{\overline{x}}(Y), \quad (\sigma , \overline{y}) \mapsto \sigma \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma ) \]

This is an action because

\[ \sigma \tau \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma \tau ) = \overline{y} \circ \mathop{\mathrm{Spec}}(\tau ) \circ \mathop{\mathrm{Spec}}(\sigma ) = \sigma \cdot (\tau \cdot \overline{y}) \]

The action is functorial in $Y \in \textit{FÉt}_ X$ and we obtain the desired map.

Comments (4)

Comment #4651 by Rex on

Shouldn't the source category of the fibre functor to be, not finite etale coverings, but finite etale coverings with basepoint?

Comment #4653 by Rex on

I'm confused. If you leave out basepoints, then objects in the source category will have nontrivial automorphisms. A natural automorphism of the fibre functor will be required to intertwine with these automorphisms. Does this not force the etale fundamental group as defined above to consist only of the central elements in what should be the fundamental group?

Comment #4655 by on

The answer to your last question is no. Everything else you say is fine (except I cannot comment on your state of mind). Suggest reading the text by Lenstra (especially section 3) to clarify things, see reference in Section 58.1.

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