The Stacks project

Lemma 97.6.1. Let $p : \mathcal{X} \to \mathcal{Y}$ and $q : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $p : \mathcal{X} \to \mathcal{Y}$ is formally smooth on objects, then so is the base change $p' : \mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Z}$ of $p$ by $q$.

Proof. This is formal. Let $U \subset U'$ be a first order thickening of affine schemes over $S$, let $z'$ be an object of $\mathcal{Z}$ over $U'$, let $w$ be an object of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $U$, and let $\delta : p'(w) \to z'|_ U$ be an isomorphism. We may write $w = (U, x, z, \alpha )$ for some object $x$ of $\mathcal{X}$ over $U$ and object $z$ of $\mathcal{Z}$ over $U$ and isomorphism $\alpha : p(x) \to q(z)$. Note that $p'(w) = z$ hence $\delta : z \to z|_ U$. Set $y' = q(z')$ and $\gamma = q(\delta ) \circ \alpha : p(x) \to y'|_ U$. As $p$ is formally smooth on objects there exists an object $x'$ of $\mathcal{X}$ over $U'$ as well as isomorphisms $\beta : x'|_ U \to x$ and $\gamma ' : p(x') \to y'$ such that (97.6.0.1) commutes. Then we consider the object $w = (U', x', z', \gamma ')$ of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $U'$ and define isomorphisms

\[ w'|_ U = (U, x'|_ U, z'|_ U, \gamma '|_ U) \xrightarrow {(\beta , \delta ^{-1})} (U, x, z, \alpha ) = w \]

and

\[ p'(w') = z' \xrightarrow {\text{id}} z'. \]

These combine to give a solution to the problem. $\square$


Comments (0)


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Your email address will not be published.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.




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 06D1. Beware of the difference between the letter 'O' and the digit '0'.