Lemma 66.18.3. In the situation of (66.18.0.1) if $Z' \to Z$ is a morphism and $z' \in |Z'|$ maps to $z$, then the induced map $F_{x, z'} \to F_{x, z}$ is surjective.

Proof. Set $W' = X \times _ Y Z' = W \times _ Z Z'$. Then $|W'| \to |W| \times _{|Z|} |Z'|$ is surjective by Properties of Spaces, Lemma 64.4.3. Hence the surjectivity of $F_{x, z'} \to F_{x, z}$. $\square$

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