Lemma 17.20.4. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module flat over $Y$. Then the functor
is exact.
Lemma 17.20.4. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module flat over $Y$. Then the functor
is exact.
Proof. This is true because $f^*\mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} = f^{-1}\mathcal{G} \otimes _{f^{-1}\mathcal{O}_ Y} \mathcal{F}$, the functor $f^{-1}$ is exact, and $\mathcal{F}$ is a flat $f^{-1}\mathcal{O}_ Y$-module. $\square$
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)