Lemma 68.18.3. In the situation of (68.18.0.1) 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 66.4.3. Hence the surjectivity of $F_{x, z'} \to F_{x, z}$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)