Example 48.3.7. The statement of Lemma 48.3.6 is not true without applying the “coherator” DQ_ Y. Indeed, suppose Y = \mathop{\mathrm{Spec}}(R) and X = \mathbf{A}^1_ R. Take L = \mathcal{O}_ X and K = \mathcal{O}_ Y. The left hand side of the arrow is in D_\mathit{QCoh}(\mathcal{O}_ Y) but the right hand side of the arrow is isomorphic to \prod _{n \geq 0} \mathcal{O}_ Y which is not quasi-coherent.
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