Corollary 11.6.0.51. The contents of this tag are now at Proposition 11.6.0.47.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Proof. Apply Proposition 7.1.5.19 in the special case $\operatorname{\mathcal{E}}= \operatorname{\mathcal{D}}$ and $\operatorname{\mathcal{E}}' = \{ D\} $. $\square$