Lemma 35.34.6. Pullback of descent data for schemes over schemes.

1. Let

$\xymatrix{ X' \ar[r]_ f \ar[d]_{a'} & X \ar[d]^ a \\ S' \ar[r]^ h & S }$

be a commutative diagram of morphisms of schemes. The construction

$(V \to X, \varphi ) \longmapsto f^*(V \to X, \varphi ) = (V' \to X', \varphi ')$

where $V' = X' \times _ X V$ and where $\varphi '$ is defined as the composition

$\xymatrix{ V' \times _{S'} X' \ar@{=}[r] & (X' \times _ X V) \times _{S'} X' \ar@{=}[r] & (X' \times _{S'} X') \times _{X \times _ S X} (V \times _ S X) \ar[d]^{\text{id} \times \varphi } \\ X' \times _{S'} V' \ar@{=}[r] & X' \times _{S'} (X' \times _ X V) & (X' \times _{S'} X') \times _{X \times _ S X} (X \times _ S V) \ar@{=}[l] }$

defines a functor from the category of descent data relative to $X \to S$ to the category of descent data relative to $X' \to S'$.

2. Given two morphisms $f_ i : X' \to X$, $i = 0, 1$ making the diagram commute the functors $f_0^*$ and $f_1^*$ are canonically isomorphic.

Proof. We omit the proof of (1), but we remark that the morphism $\varphi '$ is the morphism $(f \times f)^*\varphi$ in the notation introduced in Remark 35.34.2. For (2) we indicate which morphism $f_0^*V \to f_1^*V$ gives the functorial isomorphism. Namely, since $f_0$ and $f_1$ both fit into the commutative diagram we see there is a unique morphism $r : X' \to X \times _ S X$ with $f_ i = \text{pr}_ i \circ r$. Then we take

\begin{eqnarray*} f_0^*V & = & X' \times _{f_0, X} V \\ & = & X' \times _{\text{pr}_0 \circ r, X} V \\ & = & X' \times _{r, X \times _ S X} (X \times _ S X) \times _{\text{pr}_0, X} V \\ & \xrightarrow {\varphi } & X' \times _{r, X \times _ S X} (X \times _ S X) \times _{\text{pr}_1, X} V \\ & = & X' \times _{\text{pr}_1 \circ r, X} V \\ & = & X' \times _{f_1, X} V \\ & = & f_1^*V \end{eqnarray*}

We omit the verification that this works. $\square$

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