85.5 Morphisms of simplicial sites
We continue in the fashion described in Section 85.3 working out the meaning of morphisms of simplicial sites in cases A and B treated in that section.
In both cases we have $f'_\varphi \circ h_ n = h_ m \circ f_\varphi $ as morphisms of topoi, see Sites, Lemma 7.21.2 for case B and Sites, Definition 7.14.5 for case A.
Lemma 85.5.2. Let $\mathcal{C}_ n, f_\varphi , u_\varphi $ and $\mathcal{C}'_ n, f'_\varphi , u'_\varphi $ be as in Situation 85.3.3. Let $h$ be a morphism between simplicial sites as in Remark 85.5.1. Then we obtain a morphism of topoi
\[ h_{total} : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total}) \]
and commutative diagrams
\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n) \ar[d]_{g_ n} \ar[r]_{h_ n} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_ n) \ar[d]^{g'_ n} \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}) \ar[r]^{h_{total}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total}) } \]
Moreover, we have $(g'_ n)^{-1} \circ h_{total, *} = h_{n, *} \circ g_ n^{-1}$.
Proof.
Case A. Say $h_ n$ corresponds to the continuous functor $v_ n : \mathcal{C}'_ n \to \mathcal{C}_ n$. Then we can define a functor $v_{total} : \mathcal{C}'_{total} \to \mathcal{C}_{total}$ by using $v_ n$ in degree $n$. This is clearly a continuous functor (see definition of coverings in Lemma 85.3.1). Let $h_{total}^{-1} = v_{total, s} : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total})$ and $h_{total, *} = v_{total}^ s = v_{total}^ p : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total})$ be the adjoint pair of functors constructed and studied in Sites, Sections 7.13 and 7.14. To see that $h_{total}$ is a morphism of topoi we still have to verify that $h_{total}^{-1}$ is exact. We first observe that $(g'_ n)^{-1} \circ h_{total, *} = h_{n, *} \circ g_ n^{-1}$; this is immediate by computing sections over an object $U$ of $\mathcal{C}'_ n$. Thus, if we think of a sheaf $\mathcal{F}$ on $\mathcal{C}_{total}$ as a system $(\mathcal{F}_ n, \mathcal{F}(\varphi ))$ as in Lemma 85.3.4, then $h_{total, *}\mathcal{F}$ corresponds to the system $(h_{n, *}\mathcal{F}_ n, h_{n, *}\mathcal{F}(\varphi ))$. Clearly, the functor $(\mathcal{F}'_ n, \mathcal{F}'(\varphi )) \to (h_ n^{-1}\mathcal{F}'_ n, h_ n^{-1}\mathcal{F}'(\varphi ))$ is its left adjoint. By uniqueness of adjoints, we conclude that $h_{total}^{-1}$ is given by this rule on systems. In particular, $h_{total}^{-1}$ is exact (by the description of sheaves on $\mathcal{C}_{total}$ given in the lemma and the exactness of the functors $h_ n^{-1}$) and we have our morphism of topoi. Finally, we obtain $g_ n^{-1} \circ h_{total}^{-1} = h_ n^{-1} \circ (g'_ n)^{-1}$ as well, which proves that the displayed diagram of the lemma commutes.
Case B. Here we have a functor $v_{total} : \mathcal{C}_{total} \to \mathcal{C}'_{total}$ by using $v_ n$ in degree $n$. This is clearly a cocontinuous functor (see definition of coverings in Lemma 85.3.2). Let $h_{total}$ be the morphism of topoi associated to $v_{total}$. The commutativity of the displayed diagram of the lemma follows immediately from Sites, Lemma 7.21.2. Taking left adjoints the final equality of the lemma becomes
\[ h_{total}^{-1} \circ (g'_ n)^{Sh}_! = g^{Sh}_{n!} \circ h_ n^{-1} \]
This follows immediately from the explicit description of the functors $(g'_ n)^{Sh}_!$ and $g^{Sh}_{n!}$ in Lemma 85.3.5, the fact that $h_ n^{-1} \circ (f'_\varphi )^{-1} = f_\varphi ^{-1} \circ h_ m^{-1}$ for $\varphi : [m] \to [n]$, and the fact that we already know $h_{total}^{-1}$ commutes with restrictions to the degree $n$ parts of the simplicial sites.
$\square$
Lemma 85.5.3. With notation and hypotheses as in Lemma 85.5.2. For $K \in D(\mathcal{C}_{total})$ we have $(g'_ n)^{-1}Rh_{total, *}K = Rh_{n, *}g_ n^{-1}K$.
Proof.
Let $\mathcal{I}^\bullet $ be a K-injective complex on $\mathcal{C}_{total}$ representing $K$. Then $g_ n^{-1}K$ is represented by $g_ n^{-1}\mathcal{I}^\bullet = \mathcal{I}_ n^\bullet $ which is K-injective by Lemma 85.3.6. We have $(g'_ n)^{-1}h_{total, *}\mathcal{I}^\bullet = h_{n, *}g_ n^{-1}\mathcal{I}_ n^\bullet $ by Lemma 85.5.2 which gives the desired equality.
$\square$
In both cases we have $a'_0 \circ h_0 = h_{-1} \circ a_0$ as morphisms of topoi, see Sites, Lemma 7.21.2 for case B and Sites, Definition 7.14.5 for case A.
Lemma 85.5.5. Let $\mathcal{C}_ n, f_\varphi , u_\varphi , \mathcal{D}, a_0$, $\mathcal{C}'_ n, f'_\varphi , u'_\varphi , \mathcal{D}', a'_0$, and $h_ n$, $n \geq -1$ be as in Remark 85.5.4. Then we obtain a commutative diagram
\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}) \ar[d]_ a \ar[r]_{h_{total}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total}) \ar[d]^{a'} \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \ar[r]^{h_{-1}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}') } \]
Proof.
The morphism $h$ is defined in Lemma 85.5.2. The morphisms $a$ and $a'$ are defined in Lemma 85.4.2. Thus the only thing is to prove the commutativity of the diagram. To do this, we prove that $a^{-1} \circ h_{-1}^{-1} = h_{total}^{-1} \circ (a')^{-1}$. By the commutative diagrams of Lemma 85.5.2 and the description of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total})$ and $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_{total})$ in terms of components in Lemma 85.3.4, it suffices to show that
\[ \xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n) \ar[d]_{a_ n} \ar[r]_{h_ n} & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}'_ n) \ar[d]^{a'_ n} \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \ar[r]^{h_{-1}} & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}') } \]
commutes for all $n$. This follows from the case for $n = 0$ (which is an assumption in Remark 85.5.4) and for $n > 0$ we pick $\varphi : [0] \to [n]$ and then the required commutativity follows from the case $n = 0$ and the relations $a_ n = a_0 \circ f_\varphi $ and $a'_ n = a'_0 \circ f'_\varphi $ as well as the commutation relations $f'_\varphi \circ h_ n = h_0 \circ f_\varphi $.
$\square$
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