Lemma 96.5.6. Let $S$ be a scheme. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces as in Algebraic Stacks, Definition 93.10.1. If

1. $f$ is representable by algebraic spaces,

2. $\mathcal{Y} \to (\mathit{Sch}/S)_{fppf}$ is limit preserving on objects,

3. for an affine scheme $V$ locally of finite presentation over $S$ and $y \in \mathcal{Y}_ V$ the resulting morphism of algebraic spaces $f_ y : F_ y \to V$, see Algebraic Stacks, Equation (93.9.1.1), has property $\mathcal{P}$.

Then $f$ has property $\mathcal{P}$.

Proof. Let $V$ be a scheme over $S$ and $y \in \mathcal{Y}_ V$. We have to show that $F_ y \to V$ has property $\mathcal{P}$. Since $\mathcal{P}$ is fppf local on the base we may assume that $V$ is an affine scheme which maps into an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$. Thus we can write $V = \mathop{\mathrm{lim}}\nolimits V_ i$ with each $V_ i$ affine and of finite presentation over $\mathop{\mathrm{Spec}}(\Lambda )$, see Algebra, Lemma 10.127.2. Then $y$ comes from an object $y_ i$ over $V_ i$ for some $i$ by assumption (2). By assumption (3) the morphism $F_{y_ i} \to V_ i$ has property $\mathcal{P}$. As $\mathcal{P}$ is stable under arbitrary base change and since $F_ y = F_{y_ i} \times _{V_ i} V$ we conclude that $F_ y \to V$ has property $\mathcal{P}$ as desired. $\square$

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 07WI. Beware of the difference between the letter 'O' and the digit '0'.