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

\[ \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G}) \]

is bijective. But the right hand side is equal to

\begin{align*} \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G}) & = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, f_*f^{-1}\mathcal{G}) \\ & = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, g^{-1}f^{-1}\mathcal{G}) \\ & = \mathop{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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