The Stacks project

Lemma 29.45.4. Let $f : X \to Y$ be a morphism of schemes. If $f$ is a homeomorphism onto a closed subset of $Y$ then $f$ is affine.

Proof. Let $y \in Y$ be a point. If $y \not\in f(X)$, then there exists an affine neighbourhood of $y$ which is disjoint from $f(X)$. If $y \in f(X)$, let $x \in X$ be the unique point of $X$ mapping to $y$. Let $y \in V$ be an affine open neighbourhood. Let $U \subset X$ be an affine open neighbourhood of $x$ which maps into $V$. Since $f(U) \subset V \cap f(X)$ is open in the induced topology by our assumption on $f$ we may choose a $h \in \Gamma (V, \mathcal{O}_ Y)$ such that $y \in D(h)$ and $D(h) \cap f(X) \subset f(U)$. Denote $h' \in \Gamma (U, \mathcal{O}_ X)$ the restriction of $f^\sharp (h)$ to $U$. Then we see that $D(h') \subset U$ is equal to $f^{-1}(D(h))$. In other words, every point of $Y$ has an open neighbourhood whose inverse image is affine. Thus $f$ is affine, see Lemma 29.11.3. $\square$

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Your email address will not be published.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 04DE. Beware of the difference between the letter 'O' and the digit '0'.