Open subgroups of the arithmetic fundamental groups of algebraic curves over finite fields Let $q$ be a power of a fixed prime number $p$, and let $\mathbb{F}_q$ be the finite field of $q$ elements. For a geometrically connected algebraic curve $X$ defined over $\mathbb{F}_q$, we will denote by $\pi_1(X)$ the arithmetric fundamental group of $X$.
Now suppose that $G$ is a profinite group which is isomorphic to $\pi_1(X)$ for some smooth geometrically connected algebraic curve $X$ defined over $\mathbb{F}_q$. My question is the following.

If $H$ is an open subgroup of $G$, then is there some smooth geometrically connected algebraic curve $Y$ defined over $\mathbb{F}_q$ such that $\pi_1(Y)\cong H$?

My motivation for asking this question is that such fact holds for global number fields. That is, if $K$ is a number field, $S$ a finite set of primes of $K$ containing all all archimedean primes of $ K $ and $G_{K,S}$ the Galois group of the maximal extension of $K$ inside a fixed algebraic closure of $K$ which is unramified outside $S$, then any open subgroup of $G_{K,S}$ also has this form. In this sense, I'm asking if the analogy of the above fact for function fields holds.
 A: The answer to this question is yes, at least if you modify it slightly. Namely, for any connected locally topologically Noetherian scheme $X$, with geometric point $\overline{x}$, there is an equivalence of categories
$$\mathbf{UFEt}_X\xrightarrow{\approx}\pi_1^\mathrm{et}(X,\overline{x})\text{-}\mathbf{Set},$$
where the former category is the category of all $X$-spaces of the form $Y=\bigsqcup_i Y_i\to X$ with each $Y_i\to X$ finite etale, and where the latter category is the category of all discrete sets with a continuous action of $\pi_1^\mathrm{et}(X,\overline{x})$. This functor takes $Y\to X$ to the geometric fiber $Y_{\overline{x}}$ with its natural action of $\pi_1^\mathrm{et}(X,\overline{x})$. This equivalence induces an equivalence between the full subcategories
$$\mathbf{FEt}_X\xrightarrow{\approx}\pi_1^\mathrm{et}(X,\overline{x})\text{-}\mathbf{FinSet},$$
with the obvious meanings. Moreover, an object $Y\to X$ of $\mathbf{UFEt}_X$ is connected if and only if the associated object of $\pi_1^\mathrm{et}(X,\overline{x})\text{-}\mathbf{FinSet}$ is connected (i.e. if $\pi_1^\mathrm{et}(X,\overline{x})$ acts transitively)
In your case the open subgroup $H\subseteq G=\pi_1^\mathrm{et}(X,\overline{x})$ gives rise to the finite connected $G$-set $G/H$, and from the above this corresponds to a finite etale cover $Y\to X$. Since $X$ is smooth and projective over $\mathbb{F}_q$, so is $Y$, and is therefore also integral. Since $Y\to\mathrm{Spec}(\mathbb{F}_q)$ is projective and integral we know that $\mathcal{O}(Y)$ is a finite $\mathbb{F}_q$-domain, and thus a finite extension, say $\mathbb{F}_{q^m}$ for some $m$. Then, the natural map $Y\to \mathrm{Spec}(\mathbb{F}_{q^m})$ is a smooth, projective, geometrically connected curve.
Without the replacment of $\mathbb{F}_q$ by $\mathbb{F}_{q^m}$ this is obviously false, as you can consider from $X=\mathrm{Spec}(\mathbb{F}_q)$ and $H=G_{\mathbb{F}_{q^m}}\subseteq G_{\mathbb{F}_q}=\pi_1(X)$. In this case $Y=\mathrm{Spec}(\mathbb{F}_{q^m})\to\mathrm{Spec}(\mathbb{F}_q)$ which is evidently not geometrically connected.
