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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