Lemma 7.28.5. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a cocontinuous functor. Let $V$ be an object of $\mathcal{D}$. Let ${}^ u_ V\mathcal{I}$ be the category introduced in Section 7.19. We have a commutative diagram

$\vcenter { \xymatrix{ \, _ V^ u\mathcal{I} \ar[r]_ j \ar[d]_{u'} & \mathcal{C} \ar[d]^ u \\ \mathcal{D}/V \ar[r]^-{j_ V} & \mathcal{D} } } \quad \text{where}\quad \begin{matrix} j : (U, \psi ) \mapsto U \\ u' : (U, \psi ) \mapsto (\psi : u(U) \to V) \end{matrix}$

Declare a family of morphisms $\{ (U_ i, \psi _ i) \to (U, \psi )\}$ of ${}^ u_ V\mathcal{I}$ to be a covering if and only if $\{ U_ i \to U\}$ is a covering in $\mathcal{C}$. Then

1. ${}^ u_ V\mathcal{I}$ is a site,

2. $j$ is continuous and cocontinuous,

3. $u'$ is cocontinuous,

4. we get a commutative diagram of topoi

$\xymatrix{ \mathop{\mathit{Sh}}\nolimits ({}^ u_ V\mathcal{I}) \ar[r]_ j \ar[d]_{f'} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[d]^ f \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/V) \ar[r]^-{j_ V} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) }$

where $f$ (resp. $f'$) corresponds to $u$ (resp. $u'$), and

5. we have $f'_*j^{-1} = j_ V^{-1}f_*$.

Proof. Parts (1), (2), (3), and (4) are straightforward consequences of the definitions and the fact that the functor $j$ commutes with fibre products. We omit the details. To see (5) recall that $f_*$ is given by ${}_ su = {}_ pu$. Hence the value of $j_ V^{-1}f_*\mathcal{F}$ on $V'/V$ is the value of ${}_ pu\mathcal{F}$ on $V'$ which is the limit of the values of $\mathcal{F}$ on the category ${}^ u_{V'}\mathcal{I}$. Clearly, there is an equivalence of categories

${}^ u_{V'}\mathcal{I} \to {}^{u'}_{V'/V}\mathcal{I}$

Since the value of $f'_*j^{-1}\mathcal{F}$ on $V'/V$ is given by the limit of the values of $j^{-1}\mathcal{F}$ on the category ${}^{u'}_{V'/V}\mathcal{I}$ and since the values of $j^{-1}\mathcal{F}$ on objects of ${}^ u_ V\mathcal{I}$ are just the values of $\mathcal{F}$ (by Lemma 7.21.5 as $j$ is continuous and cocontinuous) we see that (5) is true. $\square$

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).