# Naive question in Cech cohomology

Let $$X$$ be a smooth, projective variety and $$F$$ a coherent sheaf on $$X$$. Let $$\{U_i\}_{i \in I}$$ be an open affine covering of $$X$$ and $$\{f_{ij}\}_{i with $$f_{ij} \in \Gamma(U_{ij},F)$$ satisfying the cocycle condition. Denote by $$\alpha \in H^1(X,F)$$ corresponding to the collection $$\{f_{ij}\}_{i.

Choose $$i_0 \in I$$ and a proper affine open subset $$V \subset U_{i_0}$$. Consider the new open affine covering $$\mathcal{V}$$ consisting of $$V$$ and $$U_i$$ for $$i \in I$$. We know that $$H^1(X,F)$$ does not depend on the choice of the covering. What is the Cech cocyle corresponding to $$\alpha$$ under the new open covering $$\mathcal{V}$$?

Any hint/reference will be most welcome.

• The general discussion is Tag 09UY. You should be able to work it out in this simple case. – R. van Dobben de Bruyn Oct 11 at 19:42
• @R.vanDobbendeBruyn I tried that but the notation is confusing in the reference. In this case 2 indices in $I$ maps to the same element in $J$. Could you please write down a small sketch in the case cardinality of $I$ is $3$. – Jana Oct 11 at 19:51

Let $$U = \coprod_i U_i$$ and $$U'=V \amalg \coprod_i U_i$$. Since there is $$V \to U_{i_0}$$, there is a map $$U' \to U$$ over $$X$$, and hence a map $$U'\times_X U'\to U\times_X U$$ (also over $$X$$). The original cocycle is a map $$U\times_X U\to F$$, and then the new cocycle is the induced map $$U'\times_X U' \to U\times_X U \to F$$.
To unpack that, it means that on $$V\cap U_j$$ the new cocycle is defined as the restriction of $$f_{i_0j}$$ along $$V\cap U_j \hookrightarrow U_{i_0}\cap U_j$$, and on $$U_i\cap V$$ it is defined as the restriction of $$f_{ii_0}$$ along $$U_i\cap V \hookrightarrow U_i\cap U_{i_0}$$
• My confusion is that if $i_0=j$ then there is no $f_{i_0,j}$ or am I supposed to take such $f_{i_0,j}$ to be zero? – Jana Oct 12 at 7:15
• You can define $f_{ji}=-f_{ij}$ and $f_{ii}=0$ and then ignore the ordering, and then do what I did in my answer. – David Roberts Oct 12 at 9:58