Lifting Property for Smooth Morphism

Question

My question refers to some steps in the proof of Theorem 8.8 (Lifting Property) in Bosch's "Algebraic Geometry and Commutative Algebra" (pages 380-382). Here the excerpt with red tagged unclear steps:

My point of interest is the case for smooth morphism (working with definition at page 374):

In the first part of the proof the lifting property is shown for "local" case where $X$ and $S$ were affine and now one hase to show this property for general case using finite basic affine open covering $(Y_i)_{i \in I}$ of $Y$ lying over affine subsets of $X$ and $S$ such that each $\overline{\varphi} \vert _{Y_i \cap \overline{Y}}$ can be lifted to $\varphi_i ':Y_i \to X$. The problem is to glue the lifts. Since for smooth morphisms the lifts aren't unique one can't expect that $\varphi_i ', \varphi_j '$ coincide on $Y_i \cap Y_j$ therefore the has to be modified.

According to 8.1/8 the liftings $\varphi_i ' \vert _{Y_i \cap Y_j}$ and $\varphi_j ' \vert _{Y_i \cap Y_j}$ differ by an $R$-derivation $A \to \mathfrak{b}_{ij}$.

By considerations in the excerpt the cochain $(\mathfrak{b}_{ij})_{i,j \in I}$ is an element of $\prod _{i,j \in I} Hom_{B_{ij}/\mathfrak{b}_{ij}}(\Omega_{A/R} \otimes _A(B_{ij}/\mathfrak{b}_{ij}),\mathfrak{b}_{ij})=\check C^1(\overline {\mathcal Y},\underline{Hom}_{\mathcal{O}_{\overline{Y}}}(\overline{\varphi}^*\Omega_{X/S}^1,\mathcal{J}))$ in terms of Cech cohomology with covering $\overline {\mathcal Y}:=(Y_i \cap \overline{Y})_{i \in I}$.

Now my questions:

1. Why is the cochain $(\mathfrak{b}_{ij})_{i,j \in I}$ a cocycle; so why it is mapped to zero by the map $d^1:\check C^1(\overline {\mathcal Y},\underline{Hom}_{\mathcal{O}_{\overline{Y}}}(\overline{\varphi}^*\Omega_{X/S}^1,\mathcal{J})) \to \check C^2(\overline {\mathcal Y},\underline{Hom}_{\mathcal{O}_{\overline{Y}}}(\overline{\varphi}^*\Omega_{X/S}^1,\mathcal{J}))$? I think that before we stated that the liftings don't coincide on intersections. Why should that hold for each $\mathfrak{b}_{ij}, \mathfrak{b}_{ik}$ on $Y_i \cap Y_j \cap Y_k \cap \overline{Y}$?

2. After having settled part 1. we know since 1. cohomology vanishes that $(\mathfrak{b}_{ij})_{i,j \in I}$ comes from a cochain in $\check C^0(\overline {\mathcal Y},\underline{Hom}_{\mathcal{O}_{\overline{Y}}}(\overline{\varphi}^*\Omega_{X/S}^1,\mathcal{J}))$. But how we can modify the liftings $\varphi_i '$ with help of this derived fact to make the "differences" $(\mathfrak{b}_{ij})_{i,j \in I}$ on the intersections $Y_i \cap Y_j$ vanish? Don't we get another new different summands for the $\varphi_i '$ if we kill all $\mathfrak{b}_{ij}$ such that the problem persists? Can anybody explain this argument to me?

Answer 1

1. The cocycle condition does not demand that the lifts coincide on intersections. Indeed, if $Y_{ij} := Y_i\cap Y_j,$ and $$ \epsilon_{ij} := \left.\varphi_i\right|_{Y_{ij}} - \left.\varphi_j\right|_{Y_{ij}}\in\underline{\mathrm{Hom}}_{\mathcal{O}_{\overline{Y}}}\left(\overline{\varphi}^\ast\Omega^1_{X/S},\mathcal{J}\right)(Y_{ij}) $$ then the cocycle you want is in fact the collection $$ (\epsilon_{ij})_{i,j}\in\prod_{i<j}\underline{\mathrm{Hom}}_{\mathcal{O}_{\overline{Y}}}\left(\overline{\varphi}^\ast\Omega^1_{X/S},\mathcal{J}\right)(Y_{ij}) = \check{C}^1\left(\overline{\mathcal{Y}},\underline{\mathrm{Hom}}_{\mathcal{O}_{\overline{Y}}}\left(\overline{\varphi}^\ast\Omega^1_{X/S},\mathcal{J}\right)\right), $$ (as opposed to the collection of ideals $(\mathfrak{b}_{ij})_{i,j}).$ (I hope you'll excuse my equating $Y$ and $\overline{Y}$ here, as they are the same underlying topological space.) Now the fact that this is a cocycle comes down to checking the relation $\left.\epsilon_{ij}\right|_{Y_{ijk}} + \left.\epsilon_{jk}\right|_{Y_{ijk}} = \left.\epsilon_{ik}\right|_{Y_{ijk}},$ but this is clear.
2. Now that you know the obstruction to lifting $\overline{\varphi}$ is the nonvanishing of the class $(\epsilon_{ij})_{i,j}$ as above, you can explicitly write what vanishing means to see exactly where the lift comes from. To say that $[(\epsilon_{ij})] = 0$ in $\check{\mathrm{H}}^1\left(\overline{\mathcal{Y}},\underline{\mathrm{Hom}}_{\mathcal{O}_{\overline{Y}}}\left(\overline{\varphi}^\ast\Omega^1_{X/S},\mathcal{J}\right)\right)$ means that the cocycle is in fact a coboundary; i.e., there exists $(\alpha_i)_i\in\prod_i \underline{\mathrm{Hom}}_{\mathcal{O}_{\overline{Y}}}\left(\overline{\varphi}^\ast\Omega^1_{X/S},\mathcal{J}\right)(Y_i)$ such that for all $i<j$, $$ \left.\varphi_i'\right|_{Y_{ij}} - \left.\varphi_j'\right|_{Y_{ij}} = \epsilon_{ij} = d^0((\alpha_i))_{ij} = \left.\alpha_i\right|_{Y_{ij}} - \left.\alpha_j\right|_{Y_{ij}}. $$ Set $\varphi_i := \varphi_i' - \alpha_i.$ Then by construction, we have $$ \left.\varphi_i\right|_{U_{ij}} = \left.\varphi_j\right|_{U_{ij}} $$ for all $i,j$, and hence the $\varphi_i$ glue to $\varphi$ lifting $\overline{\varphi}.$