Lemma 7.21.8. Let \mathcal{C} and \mathcal{D} be sites. Let u : \mathcal{C} \to \mathcal{D} be a functor. Assume that
u is cocontinuous,
u is continuous,
u is fully faithful,
fibre products exist in \mathcal{C} and u commutes with them, and
there exist final objects e_\mathcal {C} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}), e_\mathcal {D} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}) such that u(e_\mathcal {C}) = e_\mathcal {D}.
Let g_!, g^{-1}, g_* be as above. Then, u defines a morphism of sites f : \mathcal{D} \to \mathcal{C} with f_* = g^{-1}, f^{-1} = g_!. The composition
\xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[r]^ g & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \ar[r]^ f & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) }
is isomorphic to the identity morphism of the topos \mathop{\mathit{Sh}}\nolimits (\mathcal{C}). Moreover, the functor f^{-1} is fully faithful.
Proof.
By assumption the functor u satisfies the hypotheses of Proposition 7.14.7. Hence u defines a morphism of sites and hence a morphism of topoi f as in Lemma 7.15.2. The formulas f_* = g^{-1} and f^{-1} = g_! are clear from the lemma cited and Lemma 7.21.5. We have f_* \circ g_* = g^{-1} \circ g_* \cong \text{id}, and g^{-1} \circ f^{-1} = g^{-1} \circ g_! \cong \text{id} by Lemma 7.21.7.
We still have to show that f^{-1} is fully faithful. Let \mathcal{F}, \mathcal{G} \in \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C})). We have to show that the map
\mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G})
is bijective. But the right hand side is equal to
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G}) & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, f_*f^{-1}\mathcal{G}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, g^{-1}f^{-1}\mathcal{G}) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}) \end{align*}
(the first equality by adjunction) which proves what we want.
\square
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