Lemma 85.12.4. In Situation 85.3.3. The category of cartesian sheaves of sets (resp. abelian groups) is equivalent to the category of pairs $(\mathcal{F}, \alpha )$ where $\mathcal{F}$ is a sheaf of sets (resp. abelian groups) on $\mathcal{C}_0$ and

$\alpha : (f_{\delta _1^1})^{-1}\mathcal{F} \longrightarrow (f_{\delta _0^1})^{-1}\mathcal{F}$

is an isomorphism of sheaves of sets (resp. abelian groups) on $\mathcal{C}_1$ such that $(f_{\delta ^2_1})^{-1}\alpha = (f_{\delta ^2_0})^{-1}\alpha \circ (f_{\delta ^2_2})^{-1}\alpha$ as maps of sheaves on $\mathcal{C}_2$.

Proof. We abbreviate $d^ n_ j = f_{\delta ^ n_ j} : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{n - 1})$. The condition on $\alpha$ in the statement of the lemma makes sense because

$d^1_1 \circ d^2_2 = d^1_1 \circ d^2_1, \quad d^1_1 \circ d^2_0 = d^1_0 \circ d^2_2, \quad d^1_0 \circ d^2_0 = d^1_0 \circ d^2_1$

as morphisms of topoi $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_2) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_0)$, see Simplicial, Remark 14.3.3. Hence we can picture these maps as follows

$\xymatrix{ & (d^2_0)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[r]_-{(d^2_0)^{-1}\alpha } & (d^2_0)^{-1}(d^1_0)^{-1}\mathcal{F} \ar@{=}[rd] & \\ (d^2_2)^{-1}(d^1_0)^{-1}\mathcal{F} \ar@{=}[ru] & & & (d^2_1)^{-1}(d^1_0)^{-1}\mathcal{F} \\ & (d^2_2)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[lu]^{(d^2_2)^{-1}\alpha } \ar@{=}[r] & (d^2_1)^{-1}(d^1_1)^{-1}\mathcal{F} \ar[ru]_{(d^2_1)^{-1}\alpha } }$

and the condition signifies the diagram is commutative. It is clear that given a cartesian sheaf $\mathcal{G}$ of sets (resp. abelian groups) on $\mathcal{C}_{total}$ we can set $\mathcal{F} = \mathcal{G}_0$ and $\alpha$ equal to the composition

$(d_1^1)^{-1}\mathcal{G}_0 \to \mathcal{G}_1 \leftarrow (d_1^0)^{-1}\mathcal{G}_0$

where the arrows are invertible as $\mathcal{G}$ is cartesian. To prove this functor is an equivalence we construct a quasi-inverse. The construction of the quasi-inverse is analogous to the construction discussed in Descent, Section 35.3 from which we borrow the notation $\tau ^ n_ i : [0] \to [n]$, $0 \mapsto i$ and $\tau ^ n_{ij} : [1] \to [n]$, $0 \mapsto i$, $1 \mapsto j$. Namely, given a pair $(\mathcal{F}, \alpha )$ as in the lemma we set $\mathcal{G}_ n = (f_{\tau ^ n_ n})^{-1}\mathcal{F}$. Given $\varphi : [n] \to [m]$ we define $\mathcal{G}(\varphi ) : (f_\varphi )^{-1}\mathcal{G}_ n \to \mathcal{G}_ m$ using

$\xymatrix{ (f_\varphi )^{-1}\mathcal{G}_ n \ar@{=}[r] & (f_\varphi )^{-1}(f_{\tau ^ n_ n})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)}})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)m}})^{-1}(d^1_1)^{-1}\mathcal{F} \ar[d]^{(f_{\tau ^ m_{\varphi (n)m}})^{-1}\alpha } \\ & \mathcal{G}_ m \ar@{=}[r] & (f_{\tau ^ m_ m})^{-1}\mathcal{F} \ar@{=}[r] & (f_{\tau ^ m_{\varphi (n)m}})^{-1}(d^1_0)^{-1}\mathcal{F} }$

We omit the verification that the commutativity of the displayed diagram above implies the maps compose correctly and hence give rise to a sheaf on $\mathcal{C}_{total}$, see Lemma 85.3.4. We also omit the verification that the two functors are quasi-inverse to each other. $\square$

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