Lemma 39.14.2. Let $S$ be a scheme, let $(U, R, s, t, c)$ be a groupoid scheme over $S$. If $(\mathcal{F}, \alpha )$ is a quasi-coherent module on $(U, R, s, t, c)$ then $\alpha $ is an isomorphism.
Proof. Pull back the commutative diagram of Definition 39.14.1 by the morphism $(i, 1) : R \to R \times _{s, U, t} R$. Then we see that $i^*\alpha \circ \alpha = s^*e^*\alpha $. Pulling back by the morphism $(1, i)$ we obtain the relation $\alpha \circ i^*\alpha = t^*e^*\alpha $. By the second assumption these morphisms are the identity. Hence $i^*\alpha $ is an inverse of $\alpha $. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: