## 85.27 Descent in terms of simplicial schemes

Cartesian morphisms are defined as follows.

Definition 85.27.1. Let $a : Y \to X$ be a morphism of simplicial schemes. We say $a$ is *cartesian*, or that *$Y$ is cartesian over $X$*, if for every morphism $\varphi : [n] \to [m]$ of $\Delta $ the corresponding diagram

\[ \xymatrix{ Y_ m \ar[r]_ a \ar[d]_{Y(\varphi )} & X_ m \ar[d]^{X(\varphi )}\\ Y_ n \ar[r]^{a} & X_ n } \]

is a fibre square in the category of schemes.

Cartesian morphisms are related to descent data. First we prove a general lemma describing the category of cartesian simplicial schemes over a fixed simplicial scheme. In this lemma we denote $f^* : \mathit{Sch}/X \to \mathit{Sch}/Y$ the base change functor associated to a morphism of schemes $f :Y \to X$.

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 $\tau ^ n_ i : [0] \to [n]$, $0 \mapsto i$ 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 $Y_ n = X_ n \times _{X(\tau ^ n_ n), X_0} V$. 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.
$\square$

Definition 85.27.3. Let $f : X \to S$ be a morphism of schemes. The *simplicial scheme associated to $f$*, denoted $(X/S)_\bullet $, is the functor $\Delta ^{opp} \to \mathit{Sch}$, $[n] \mapsto X \times _ S \ldots \times _ S X$ described in Simplicial, Example 14.3.5.

Thus $(X/S)_ n$ is the $(n + 1)$-fold fibre product of $X$ over $S$. The morphism $d^1_0 : X \times _ S X \to X$ is the map $(x_0, x_1) \mapsto x_1$ and the morphism $d^1_1$ is the other projection. The morphism $s^0_0$ is the diagonal morphism $X \to X \times _ S X$.

Lemma 85.27.4. Let $f : X \to S$ be a morphism of schemes. Let $\pi : Y \to (X/S)_\bullet $ be a cartesian morphism of simplicial schemes. Set $V = Y_0$ considered as a scheme over $X$. The morphisms $d^1_0, d^1_1 : Y_1 \to Y_0$ and the morphism $\pi _1 : Y_1 \to X \times _ S X$ induce isomorphisms

\[ \xymatrix{ V \times _ S X & & Y_1 \ar[ll]_-{(d^1_1, \text{pr}_1 \circ \pi _1)} \ar[rr]^-{(\text{pr}_0 \circ \pi _1, d^1_0)} & & X \times _ S V. } \]

Denote $\varphi : V \times _ S X \to X \times _ S V$ the resulting isomorphism. Then the pair $(V, \varphi )$ is a descent datum relative to $X \to S$.

**Proof.**
This is a special case of (part of) Lemma 85.27.2 as the displayed equation of that lemma is equivalent to the cocycle condition of Descent, Definition 35.34.1.
$\square$

Lemma 85.27.5. Let $f : X \to S$ be a morphism of schemes. The construction

\[ \begin{matrix} \text{category of cartesian }
\\ \text{schemes over } (X/S)_\bullet
\end{matrix} \longrightarrow \begin{matrix} \text{ category of descent data}
\\ \text{ relative to } X/S
\end{matrix} \]

of Lemma 85.27.4 is an equivalence of categories.

**Proof.**
The functor from left to right is given in Lemma 85.27.4. Hence this is a special case of Lemma 85.27.2.
$\square$

We may reinterpret the pullback of Descent, Lemma 35.34.6 as follows. Suppose given a morphism of simplicial schemes $f : X' \to X$ and a cartesian morphism of simplicial schemes $Y \to X$. Then the fibre product (viewed as a “pullback”)

\[ f^*Y = Y \times _ X X' \]

of simplicial schemes is a simplicial scheme cartesian over $X'$. Suppose given a commutative diagram of morphisms of schemes

\[ \xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S. } \]

This gives rise to a morphism of simplicial schemes

\[ f_\bullet : (X'/S')_\bullet \longrightarrow (X/S)_\bullet . \]

We claim that the “pullback” $f_\bullet ^*$ along the morphism $f_\bullet : (X'/S')_\bullet \to (X/S)_\bullet $ corresponds via Lemma 85.27.5 with the pullback defined in terms of descent data in the aforementioned Descent, Lemma 35.34.6.

## Comments (0)