Lemma 37.34.8. Let $f : X \to S$ be a morphism of affine schemes, which is of finite presentation with geometrically connected fibres. Then there exists a diagram as in Lemma 37.34.1 such that in addition $f_0$ has geometrically connected fibres.
Proof. Apply Lemma 37.34.1 to get a cartesian diagram
\[ \xymatrix{ X_0 \ar[d]_{f_0} & X \ar[l]^ g \ar[d]^ f \\ S_0 & S \ar[l]_ h } \]
of affine schemes with $X_0 \to S_0$ a finite type morphism of schemes of finite type over $\mathbf{Z}$. By Lemma 37.28.6 the set $E \subset S_0$ of points where the fibre of $f_0$ is geometrically connected is a constructible subset. By Lemma 37.28.2 we have $h(S) \subset E$. Write $S_0 = \mathop{\mathrm{Spec}}(A_0)$ and $S = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$ as a direct colimit of finite type $A_0$-algebras. By Limits, Lemma 32.4.10 we see that $\mathop{\mathrm{Spec}}(A_ i) \to S_0$ has image contained in $E$ for some $i$. After replacing $S_0$ by $\mathop{\mathrm{Spec}}(A_ i)$ and $X_0$ by $X_0 \times _{S_0} \mathop{\mathrm{Spec}}(A_ i)$ we see that all fibres of $f_0$ are geometrically connected. $\square$
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