Algebraic fundamental group of $\mathbb{P}^1(\mathbb{Q})\setminus\{P_1,\ldots,P_n\}$

Question

Define the algebraic fundamental group as follows:

Let $k$ be a perfect field, $X$ an integral proper normal $k$-curve with function field $K$ and $U\subset X$ a nonempty open subset. Denote by $K_s$ a fixed separable closure of $K$.Let $K_U$ be the composite of all finite subextensions $L:K_s/L/K$ such that the corresponding finite morphism of proper normal curves is étale over $U$. Then $K_U$ is Galois over $K$ and the algebraic fundamental group of $U$ is $\operatorname{Gal}(K_U/K)$, denoted as $\pi_1(U)$.
(Cf. Tamas Szamuely, Galois Group and Fundamental Groups, Section 4.6)

Now let $\Pi(n)$ be $\pi_1(\mathbb{P}^1(\mathbb{Q})\setminus\{P_1,\ldots,P_n\})$. Then why is $\Pi(n)$ a quotient of $\operatorname{Gal}\bigl(\overline{\mathbb{Q}(t)}/\mathbb{Q}(t)\bigr)$?

(Section 4.8 of the book I mentioned above)

Answer 1

The function field of $X = \mathbf{P}^1(k)\setminus\{P_1,\ldots,P_n\}$ is $k(t)$. As you described, a finite étale cover of $X$ corresponds to an extension $L/k(t)$ in a fixed separable closure of $k(t)$. The compositum $K_U/k(t)$ of all such extensions is Galois and the étale fundamental group of $X$ is defined as the Galois group of $K_U/k(t)$.

To answer your question recall that if an intermediate field $M/K$ of a Galois extension $L/K$ is Galois, then

$$\text{Gal}(M/K) \cong \text{Gal}(L/K)/\text{Gal}(L/M)$$

In particular,

$$\pi_1^{et}(\mathbf{P}^1(k)\setminus\{P_1,\ldots,P_n\}) = \text{Gal}(K_U/k(t))$$

is a quotient of $\text{Gal}(k(t)^s/k(t))$.