Collapsing a fibre of a proper family forces nearby ones to collapse too.

Lemma 70.35.3. Let $S$ be a scheme. Let

$\xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & Y \ar[ld]^ g \\ & B }$

be a commutative diagram of morphism of algebraic spaces over $S$. Let $b \in B$ and let $\mathop{\mathrm{Spec}}(k) \to B$ be a morphism in the equivalence class of $b$. Assume

1. $X \to B$ is a proper morphism,

2. $Y \to B$ is separated and locally of finite type,

3. one of the following is true

1. the image of $|X_ k| \to |Y_ k|$ is finite,

2. the image of $|f|^{-1}(\{ b\} )$ in $|Y|$ is finite and $B$ is decent.

Then there is an open subspace $B' \subset B$ containing $b$ such that $X_{B'} \to Y_{B'}$ factors through a closed subspace $Z \subset Y_{B'}$ finite over $B'$.

Proof. Let $Z \subset Y$ be the scheme theoretic image of $h$, see Morphisms of Spaces, Section 61.16. By Morphisms of Spaces, Lemma 61.40.8 the morphism $X \to Z$ is surjective and $Z \to B$ is proper. Thus

$\{ x \in |X|\text{ lying over }b\} \to \{ z \in |Z|\text{ lying over }b\}$

and $|X_ k| \to |Z_ k|$ are surjective. We see that either (3)(a) or (3)(b) imply that $Z \to B$ is quasi-finite all points of $|Z|$ lying over $b$ by Decent Spaces, Lemma 62.18.10. Hence $Z \to B$ is finite in an open neighbourhood of $b$ by Lemma 70.35.2. $\square$

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AEJ. Beware of the difference between the letter 'O' and the digit '0'.