The Stacks Project


Tag 01U1

Chapter 28: Morphisms of Schemes > Section 28.22: Open morphisms

Lemma 28.22.2. Let $f : X \to S$ be a morphism.

  1. If $f$ is locally of finite presentation and generalizations lift along $f$, then $f$ is open.
  2. If $f$ is locally of finite presentation and generalizations lift along every base change of $f$, then $f$ is universally open.

Proof. It suffices to prove the first assertion. This reduces to the case where both $X$ and $S$ are affine. In this case the result follows from Algebra, Lemma 10.40.3 and Proposition 10.40.8. $\square$

    The code snippet corresponding to this tag is a part of the file morphisms.tex and is located in lines 4034–4043 (see updates for more information).

    \begin{lemma}
    \label{lemma-locally-finite-presentation-universally-open}
    Let $f : X \to S$ be a morphism.
    \begin{enumerate}
    \item If $f$ is locally of finite presentation and generalizations lift
    along $f$, then $f$ is open.
    \item If $f$ is locally of finite presentation and generalizations lift
    along every base change of $f$, then $f$ is universally open.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    It suffices to prove the first assertion.
    This reduces to the case where both $X$ and $S$ are affine.
    In this case the result follows from
    Algebra, Lemma \ref{algebra-lemma-going-up-down-specialization}
    and Proposition \ref{algebra-proposition-fppf-open}.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 01U1

    Your email address will not be published.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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?