Processing math: 100%

The Stacks project

Lemma 7.21.8. Let \mathcal{C} and \mathcal{D} be sites. Let u : \mathcal{C} \to \mathcal{D} be a functor. Assume that

  1. u is cocontinuous,

  2. u is continuous,

  3. u is fully faithful,

  4. fibre products exist in \mathcal{C} and u commutes with them, and

  5. 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


Comments (0)


Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.