Lemma 37.11.5. Let $f : X \to S$ be a morphism of schemes. Let $U \subset X$ and $V \subset S$ be open subschemes such that $f(U) \subset V$. If $f$ is formally smooth, so is $f|_ U : U \to V$.

**Proof.**
Consider a solid diagram

as in Definition 37.11.1. If $f$ is formally smooth, then there exists an $S$-morphism $a' : T' \to X$ such that $a'|_ T = a$. Since the underlying sets of $T$ and $T'$ are the same we see that $a'$ is a morphism into $U$ (see Schemes, Section 26.3). And it clearly is a $V$-morphism as well. Hence the dotted arrow above as desired. $\square$

## 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Your email address will not be published.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)

There are also: