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Proof. Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa (s) \to \kappa (\overline{s})$. Let $A \to \kappa (\overline{s}')$ be the map used to construct the first specialization map. Let $\mathcal{O}_{S, s'} \to A'$ be the strict henselization constructed using $\kappa (s') \subset \kappa (\overline{s}')$. By functoriality of strict henselization, there is a map $A \to A'$ such that the composition with $A' \to \kappa (\overline{s}')$ is the given map (Algebra, Lemma 10.154.6). Next, let $A' \to \kappa (\overline{s}'')$ be the map used to construct the second specialization map. Then it is clear that the composition of the first and second specialization maps is the specialization map $\pi _1(X_{\overline{s}''}) \to \pi _1(X_{\overline{s}})$ constructed using $A \to A' \to \kappa (\overline{s}'')$. $\square$