Lemma 7.20.3. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. The functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$, $\mathcal{G} \mapsto (u^ p\mathcal{G})^\# $ is a left adjoint to the functor ${}_ su$ introduced in Lemma 7.20.2 above. Moreover, it is exact.
Proof. Let us prove the adjointness property as follows
\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})} ((u^ p\mathcal{G})^\# , \mathcal{F}) & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})} (u^ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{D})} (\mathcal{G}, {}_ pu\mathcal{F}) \\ & = & \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})} (\mathcal{G}, {}_ su\mathcal{F}). \end{eqnarray*}
Thus it is a left adjoint and hence right exact, see Categories, Lemma 4.24.6. We have seen that sheafification is left exact, see Lemma 7.10.14. Moreover, the inclusion $i : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \textit{PSh}(\mathcal{D})$ is left exact by Lemma 7.10.1. Finally, the functor $u^ p$ is left exact because it is a right adjoint (namely to $u_ p$). Thus the functor is the composition ${}^\# \circ u^ p \circ i$ of left exact functors, hence left exact. $\square$
Post a comment
Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: