The Stacks project

Lemma 58.16.1. Consider a commutative diagram

\[ \xymatrix{ Y \ar[d]_ g \ar[r] & X \ar[d]^ f \\ T \ar[r] & S } \]

of schemes where $f$ and $g$ are proper with geometrically connected fibres. Let $t' \leadsto t$ be a specialization of points in $T$ and consider a specialization map $sp : \pi _1(Y_{\overline{t}'}) \to \pi _1(Y_{\overline{t}})$ as above. Then there is a commutative diagram

\[ \xymatrix{ \pi _1(Y_{\overline{t}'}) \ar[r]_{sp} \ar[d] & \pi _1(Y_{\overline{t}}) \ar[d] \\ \pi _1(X_{\overline{s}'}) \ar[r]^{sp} & \pi _1(X_{\overline{s}}) } \]

of specialization maps where $\overline{s}$ and $\overline{s}'$ are the images of $\overline{t}$ and $\overline{t}'$.

Proof. Let $B$ be the strict henselization of $\mathcal{O}_{T, t}$ with respect to $\kappa (t) \subset \kappa (t)^{sep} \subset \kappa (\overline{t})$. Pick $\psi : \overline{t}' \to \mathop{\mathrm{Spec}}(B)$ lifting $\overline{t}' \to T$ as in the construction of the specialization map. Let $s$ and $s'$ denote the images of $t$ and $t'$ in $S$. Let $A$ be the strict henselization of $\mathcal{O}_{S, s}$ with respect to $\kappa (s) \subset \kappa (s)^{sep} \subset \kappa (\overline{s})$. Since $\kappa (\overline{s}) = \kappa (\overline{t})$, by the functoriality of strict henselization (Algebra, Lemma 10.155.10) we obtain a ring map $A \to B$ fitting into the commutative diagram

\[ \xymatrix{ \overline{t}' \ar[r]_-\psi \ar[d] & \mathop{\mathrm{Spec}}(B) \ar[d] \ar[r] & T \ar[d] \\ \overline{s}' \ar[r]^-\varphi & \mathop{\mathrm{Spec}}(A) \ar[r] & S } \]

Here the morphism $\varphi : \overline{s}' \to \mathop{\mathrm{Spec}}(A)$ is simply taken to be the composition $\overline{t}' \to \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Applying base change we obtain a commutative diagram

\[ \xymatrix{ Y_{\overline{t}'} \ar[r] \ar[d] & Y_ B \ar[d] \\ X_{\overline{s}'} \ar[r] & X_ A } \]

and from the construction of the specialization map the commutativity of this diagram implies the commutativity of the diagram of the lemma. $\square$

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