The Stacks project

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

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


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

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)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0248. Beware of the difference between the letter 'O' and the digit '0'.