The map is surjective by Lemma \ref{lemma-specialization-map-surjective}. Thus we have to show it is injective. \medskip\noindent We may assume $S$ is affine. Then $S$ is a cofiltered limit of affine schemes of finite type over $\mathbf{Z}$. Hence we can assume $X \to S$ is the base change of $X_0 \to S_0$ where $S_0$ is the spectrum of a finite type $\mathbf{Z}$-algebra and $X_0 \to S_0$ is smooth and proper. See Limits, Lemma \ref{limits-lemma-descend-finite-presentation}, \ref{limits-lemma-descend-smooth}, and \ref{limits-lemma-eventually-proper}. By Lemma \ref{lemma-specialization-map-base-change} we reduce to the case where the base is Noetherian. \medskip\noindent Applying Lemma \ref{lemma-specialization-map-discrete-valuation-ring} we reduce to the case where the base $S$ is the spectrum of a strictly henselian discrete valuation ring $A$ and we are looking at the specialization map over $A$. Let $K$ be the fraction field of $A$. Choose an algebraic closure $\overline{K}$ which corresponds to a geometric generic point $\overline{\eta}$ of $\Spec(A)$. For $\overline{K}/L/K$ finite separable, let $B \subset L$ be the integral closure of $A$ in $L$. This is a discrete valuation ring by More on Algebra, Remark \ref{more-algebra-remark-finite-separable-extension}. \medskip\noindent Let $X \to \Spec(A)$ be as in the previous paragraph. To show injectivity of the specialization map it suffices to prove that every finite \'etale cover $V$ of $X_{\overline{\eta}}$ is the base change of a finite \'etale cover $Y \to X$. Namely, then $\pi_1(X_{\overline{\eta}}) \to \pi_1(X) = \pi_1(X_s)$ is injective by Lemma \ref{lemma-functoriality-galois-injective}. \medskip\noindent Given $V$ we can first descend $V$ to $V' \to X_{K^{sep}}$ by Lemma \ref{lemma-perfection} and then to $V'' \to X_L$ by Lemma \ref{lemma-limit}. Let $Z \to X_B$ be the normalization of $X_B$ in $V''$. Observe that $Z$ is normal and that $Z_L = V''$ as schemes over $X_L$. Hence $Z \to X_B$ is finite \'etale over the generic fibre. The problem is that we do not know that $Z \to X_B$ is everywhere \'etale. Since $X \to \Spec(A)$ has geometrically connected smooth fibres, we see that the special fibre $X_s$ is geometrically irreducible. Hence the special fibre of $X_B \to \Spec(B)$ is irreducible; let $\xi_B$ be its generic point. Let $\xi_1, \ldots, \xi_r$ be the points of $Z$ mapping to $\xi_B$. Our first (and it will turn out only) problem is now that the extensions $$ \mathcal{O}_{X_B, \xi_B} \subset \mathcal{O}_{Z, \xi_i} $$ of discrete valuation rings may be ramified. Let $e_i$ be the ramification index of this extension. Note that since the characteristic of $\kappa(s)$ is zero, the ramification is tame! \medskip\noindent To get rid of the ramification we are going to choose a further finite separable extension $K^{sep}/L'/L/K$ such that the ramification index $e$ of the induced extensions $B'/B$ is divisible by $e_i$. Consider the normalized base change $Z'$ of $Z$ with respect to $\Spec(B') \to \Spec(B)$, see discussion in More on Morphisms, Section \ref{more-morphisms-section-reduced-fibre-theorem}. Let $\xi_{i, j}$ be the points of $Z'$ mapping to $\xi_{B'}$ and to $\xi_i$ in $Z$. Then the local rings $$ \mathcal{O}_{Z', \xi_{i, j}} $$ are localizations of the integral closure of $\mathcal{O}_{Z, \xi_i}$ in $L' \otimes_L F_i$ where $F_i$ is the fraction field of $\mathcal{O}_{Z, \xi_i}$; details omitted. Hence Abhyankar's lemma (More on Algebra, Lemma \ref{more-algebra-lemma-abhyankar}) tells us that $$ \mathcal{O}_{X_{B'}, \xi_{B'}} \subset \mathcal{O}_{Z', \xi_{i, j}} $$ is unramified. We conclude that the morphism $Z' \to X_{B'}$ is \'etale away from codimension $1$. Hence by purity of branch locus (Lemma \ref{lemma-purity}) we see that $Z' \to X_{B'}$ is finite \'etale! \medskip\noindent However, since the residue field extension induced by $A \to B'$ is trivial (as the residue field of $A$ is algebraically closed being separably closed of characteristic zero) we conclude that $Z'$ is the base change of a finite \'etale cover $Y \to X$ by applying Lemma \ref{lemma-finite-etale-on-proper-over-henselian} twice (first to get $Y$ over $A$, then to prove that the pullback to $B$ is isomorphic to $Z'$). This finishes the proof.