Lemma 83.12.9. In Situation 83.3.3.

1. An object $K$ of $D(\mathcal{C}_{total})$ is cartesian if and only the canonical map

$g_{n!}K_ n \longrightarrow g_{n!}\mathbf{Z} \otimes ^\mathbf {L}_\mathbf {Z} K$

is an isomorphism for all $n$.

2. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{C}_{total}$ such that the morphisms $f_\varphi ^{-1}\mathcal{O}_ n \to \mathcal{O}_ m$ are flat for all $\varphi : [n] \to [m]$. Then an object $K$ of $D(\mathcal{O})$ is cartesian if and only if the canonical map

$g_{n!}K_ n \longrightarrow g_{n!}\mathcal{O}_ n \otimes ^\mathbf {L}_\mathcal {O} K$

is an isomorphism for all $n$.

Proof. Proof of (1). Since $g_{n!}$ is exact, it induces a functor on derived categories adjoint to $g_ n^{-1}$. The map is the adjoint of the map $K_ n \to (g_ n^{-1}g_{n!}\mathbf{Z}) \otimes ^\mathbf {L}_\mathbf {Z} K_ n$ corresponding to $\mathbf{Z} \to g_ n^{-1}g_{n!}\mathbf{Z}$ which in turn is adjoint to $\text{id} : g_{n!}\mathbf{Z} \to g_{n!}\mathbf{Z}$. Using the description of $g_{n!}$ given in Lemma 83.3.5 we see that the restriction to $\mathcal{C}_ m$ of this map is

$\bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^{-1}K_ n \longrightarrow \bigoplus \nolimits _{\varphi : [n] \to [m]} K_ m$

Thus the statement is clear.

Proof of (2). Since $g_{n!}$ is exact (Lemma 83.6.3), it induces a functor on derived categories adjoint to $g_ n^*$ (also exact). The map is the adjoint of the map $K_ n \to (g_ n^*g_{n!}\mathcal{O}_ n) \otimes ^\mathbf {L}_{\mathcal{O}_ n} K_ n$ corresponding to $\mathcal{O}_ n \to g_ n^*g_{n!}\mathcal{O}_ n$ which in turn is adjoint to $\text{id} : g_{n!}\mathcal{O}_ n \to g_{n!}\mathcal{O}_ n$. Using the description of $g_{n!}$ given in Lemma 83.6.1 we see that the restriction to $\mathcal{C}_ m$ of this map is

$\bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*K_ n \longrightarrow \bigoplus \nolimits _{\varphi : [n] \to [m]} f_\varphi ^*\mathcal{O}_ n \otimes _{\mathcal{O}_ m} K_ m = \bigoplus \nolimits _{\varphi : [n] \to [m]} K_ m$

Thus the statement is clear. $\square$

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