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Lemma 31.36.2. Let $X_\bullet$ be a simplicial scheme. The category of simplicial schemes cartesian over $X_\bullet$ 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 13.3.3 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 $V_\bullet$ cartesian over $X_\bullet$ we can set $V = V_0$ and $\varphi$ equal to the composition $$ V \times_{X_0, d^1_1} X_1 = V_1 = X_1 \times_{X_0, d^1_0} V $$ 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 Section 31.3 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 $(V, \varphi)$ as in the lemma we set $V_n = X_n \times_{X(\tau^n_n), X_0} V$. Then given $\beta : [n] \to [m]$ we define $V(\beta) : V_m \to V_n$ as the pullback by $X(\tau^m_{\beta(n)m})$ of the map $\varphi$ postcomposed by the projection $X_m \times_{X(\beta), X_n} V_n \to V_n$. This makes sense because $$ X_m \times_{X(\tau^m_{\beta(n)m}), X_1} X_1 \times_{d^1_1, X_0} V = X_m \times_{X(\tau^m_m), X_0} V = V_m $$ and $$ X_m \times_{X(\tau^m_{\beta(n)m}), X_1} X_1 \times_{d^1_0, X_0} V = X_m \times_{X(\tau^m_{\beta(n)}), X_0} V = X_m \times_{X(\beta), X_n} V_n. $$ 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. $\square$

\begin{lemma}
\label{lemma-characterize-cartesian-schemes}
Let $X_\bullet$ be a simplicial scheme.
The category of simplicial schemes cartesian over $X_\bullet$
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 $X_2$.
\end{lemma}

\begin{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 \ref{simplicial-remark-relations} 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 $V_\bullet$ cartesian over $X_\bullet$ we can
set $V = V_0$ and $\varphi$ equal to the composition
$$
V \times_{X_0, d^1_1} X_1 = V_1 = X_1 \times_{X_0, d^1_0} V
$$
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
Section \ref{section-descent-modules} 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 $(V, \varphi)$
as in the lemma we set $V_n = X_n \times_{X(\tau^n_n), X_0} V$.
Then given $\beta : [n] \to [m]$ we define
$V(\beta) : V_m \to V_n$ as the pullback by $X(\tau^m_{\beta(n)m})$
of the map $\varphi$ postcomposed by the projection
$X_m \times_{X(\beta), X_n} V_n \to V_n$. This makes sense because
$$
X_m \times_{X(\tau^m_{\beta(n)m}), X_1} X_1 \times_{d^1_1, X_0} V
=
X_m \times_{X(\tau^m_m), X_0} V = V_m
$$
and
$$
X_m \times_{X(\tau^m_{\beta(n)m}), X_1} X_1 \times_{d^1_0, X_0} V =
X_m \times_{X(\tau^m_{\beta(n)}), X_0} V =
X_m \times_{X(\beta), X_n} V_n.
$$
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.
\end{proof}

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