Remark 18.40.2. Warning! Let $u : \mathcal{C} \to \mathcal{D}$, $g$, $\mathcal{O}_\mathcal {D}$, and $\mathcal{O}_\mathcal {C}$ be as in Lemma 18.40.1. In general it is **not** the case that the diagram

commutes (here $g^{Ab}_!$ is the one from Lemma 18.16.2). There is a transformation of functors

From the proof of Lemma 18.40.1 we see that this is an isomorphism if and only if $g^{Ab}_!j_{U!}\mathcal{O}_ U \to g_!j_{U!}\mathcal{O}_ U$ is an isomorphism for all objects $U$ of $\mathcal{C}$. Since we have $g_!j_{U!}\mathcal{O}_ U = j_{u(U)!}\mathcal{O}_{u(U)}$ this holds if and only if

is an isomorphism for all objects $U$ of $\mathcal{C}$. Note that for such a $U$ we obtain a commutative diagram

of cocontinuous functors of sites, see Sites, Lemma 7.28.4 and therefore $g^{Ab}_!j_{U!} = j_{u(U)!}(g')^{Ab}_!$ where $g' : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/u(U))$ is the morphism of topoi induced by the cocontinuous functor $u'$. Hence we see that $g_! = g^{Ab}_!$ if the canonical map

is an isomorphism for all objects $U$ of $\mathcal{C}$.

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