# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Go back to the page of Proposition 6.3.1.19.

Comment #477 by Haoqing on

I guess the target in "$\theta : \operatorname{Fun}(\operatorname{\mathcal{D}}, \operatorname{\mathcal{E}})^{\simeq } \rightarrow \operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})$ " should be $\operatorname{Fun}( \operatorname{\mathcal{C}}[W^{-1}], \operatorname{\mathcal{E}})^{\simeq}$ which is again equivalent to $\operatorname{Fun}( \operatorname{\mathcal{C}}, \operatorname{\mathcal{E}^{\simeq}})$?

Comment #480 by Kerodon on

Yep. Thanks!

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