The Stacks project

Lemma 20.49.4. Let $f : X \to Y$ be a morphism of ringed spaces such that $f$ is a homeomorphism onto a closed subset. Then (20.49.2.1) is an isomorphism always.

Proof. Since $f$ is a homeomorphism onto a closed subset, the functor $f_*$ is exact (Modules, Lemma 17.6.1). Hence $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a K-flat complex $\mathcal{K}^\bullet $ of $\mathcal{O}_ Y$-modules representing $K$ and choose any complex $\mathcal{E}^\bullet $ of $\mathcal{O}_ X$-modules representing $E$. Then $Lf^*K$ is represented by $f^*\mathcal{K}^\bullet $ which is a K-flat complex of $\mathcal{O}_ X$-modules (Lemma 20.26.7). Thus the right hand side of (20.49.2.1) is represented by

\[ f_*\text{Tot}(\mathcal{E}^\bullet \otimes _{\mathcal{O}_ X} f^*\mathcal{K}^\bullet ) \]

By the same reasoning we see that the left hand side is represented by

\[ \text{Tot}(f_*\mathcal{E}^\bullet \otimes _{\mathcal{O}_ Y} \mathcal{K}^\bullet ) \]

Since $f_*$ commutes with direct sums (Modules, Lemma 17.6.3) it suffices to show that

\[ f_*(\mathcal{E} \otimes _{\mathcal{O}_ X} f^*\mathcal{K}) = f_*\mathcal{E} \otimes _{\mathcal{O}_ Y} \mathcal{K} \]

for any $\mathcal{O}_ X$-module $\mathcal{E}$ and $\mathcal{O}_ Y$-module $\mathcal{K}$. We will check this by checking on stalks. Let $y \in Y$. If $y \not\in f(X)$, then the stalks of both sides are zero. If $y = f(x)$, then we see that we have to show

\[ \mathcal{E}_ x \otimes _{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y) = \mathcal{E}_ x \otimes _{\mathcal{O}_{Y, y}} \mathcal{F}_ y \]

(using Sheaves, Lemma 6.32.1 and Lemma 6.26.4). This equality holds and therefore the lemma has been proved. $\square$


Comments (2)

Comment #1484 by jojo on

In the statement I think it should say that is a homeomorphism onto a closed subset (not a homomorphism).


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Your email address will not be published.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 0B55. Beware of the difference between the letter 'O' and the digit '0'.