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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