Lemma 85.27.2. Let $X$ be a simplicial scheme. The category of simplicial schemes cartesian over $X$ is equivalent to the category of pairs $(V, \varphi )$ where $V$ is a scheme over $X_0$ and

$\varphi : V \times _{X_0, d^1_1} X_1 \longrightarrow X_1 \times _{d^1_0, X_0} V$

is an isomorphism over $X_1$ such that $(s_0^0)^*\varphi = \text{id}_ V$ and such that

$(d^2_1)^*\varphi = (d^2_0)^*\varphi \circ (d^2_2)^*\varphi$

as morphisms of schemes over $X_2$.

Proof. The statement of the displayed equality makes sense because $d^1_1 \circ d^2_2 = d^1_1 \circ d^2_1$, $d^1_1 \circ d^2_0 = d^1_0 \circ d^2_2$, and $d^1_0 \circ d^2_0 = d^1_0 \circ d^2_1$ as morphisms $X_2 \to X_0$, see Simplicial, Remark 14.3.3 hence we can picture these maps as follows

$\xymatrix{ & X_2 \times _{d^1_1 \circ d^2_0, X_0} V \ar[r]_-{(d^2_0)^*\varphi } & X_2 \times _{d^1_0 \circ d^2_0, X_0} V \ar@{=}[rd] & \\ X_2 \times _{d^1_0 \circ d^2_2, X_0} V \ar@{=}[ru] & & & X_2 \times _{d^1_0 \circ d^2_1, X_0} V \\ & X_2 \times _{d^1_1 \circ d^2_2, X_0} V \ar[lu]^{(d^2_2)^*\varphi } \ar@{=}[r] & X_2 \times _{d^1_1 \circ d^2_1, X_0} V \ar[ru]_{(d^2_1)^*\varphi } }$

and the condition signifies the diagram is commutative. It is clear that given a simplicial scheme $Y$ cartesian over $X$ we can set $V = Y_0$ and $\varphi$ equal to the composition

$V \times _{X_0, d^1_1} X_1 = Y_0 \times _{X_0, d^1_1} X_1 = Y_1 = X_1 \times _{X_0, d^1_0} Y_0 = X_1 \times _{X_0, d^1_0} V$

of identifications given by the cartesian structure. 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 $\tau ^ n_ i :  \to [n]$, $0 \mapsto i$ and $\tau ^ n_{ij} :  \to [n]$, $0 \mapsto i$, $1 \mapsto j$. Namely, given a pair $(V, \varphi )$ as in the lemma we set $Y_ n = X_ n \times _{X(\tau ^ n_ n), X_0} V$. Then given $\beta : [n] \to [m]$ we define $V(\beta ) : Y_ m \to Y_ n$ as the pullback by $X(\tau ^ m_{\beta (n)m})$ of the map $\varphi$ postcomposed by the projection $X_ m \times _{X(\beta ), X_ n} Y_ n \to Y_ n$. This makes sense because

$X_ m \times _{X(\tau ^ m_{\beta (n)m}), X_1} X_1 \times _{d^1_1, X_0} V = X_ m \times _{X(\tau ^ m_ m), X_0} V = Y_ m$

and

$X_ m \times _{X(\tau ^ m_{\beta (n)m}), X_1} X_1 \times _{d^1_0, X_0} V = X_ m \times _{X(\tau ^ m_{\beta (n)}), X_0} V = X_ m \times _{X(\beta ), X_ n} Y_ n.$

We omit the verification that the commutativity of the displayed diagram above implies the maps compose correctly. We also omit the verification that the two functors are quasi-inverse to each other. $\square$

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