Lemma 74.22.6. Pullback of descent data. Let $S$ be a scheme.

1. Let

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

be a commutative diagram of algebraic spaces over $S$. The construction

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

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

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

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

2. Given two morphisms $f_ i : Y' \to Y$, $i = 0, 1$ 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 74.22.2. 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 $r : Y' \to Y \times _ X Y$ with $f_ i = \text{pr}_ i \circ r$. Then we take

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

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

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).