# Cup product of Cech cohomology graded commutative

Let $(X,\mathscr{O}_X)$ be a ringed space. Let $\{U_i\}_{i\in I}$ be a finite open cover of $X$. Let $U_{i_0\cdots i_n}=\cap_{i_j}U_{i_j}$. Consider the Cech cohomology given by the this cover, $\check{H}^j(\{U_i\},\mathscr{O}_X)$.

For general $\mathscr{O}_X$ modules $\mathscr{F}$ and $\mathscr{G}$, there is a cup product of $$\check{H}^p(\{U_i\},\mathscr{F})\times \check{H}^q(\{U_i\},\mathscr{G})\to \check{H}^{p+q}(\{U_i\},\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G})$$

taking $s\in C^p(\{U_i\}, \mathscr{F})$ and $t\in C^q(\{U_i\}, \mathscr{G})$ to $s\otimes t$ given on $U_{i_0\cdots i_{p_q}}$ by $s_{i_0\cdots i_p}\otimes t_{i_p\cdots i_{p+q}}$. If $\mathscr{F},\mathscr{G}=\mathscr{O}_X$, we have

$$\smile:\check{H}^p(\{U_i\},\mathscr{O}_X)\times \check{H}^q(\{U_i\},\mathscr{O}_X)\to \check{H}^{p+q}(\{U_i\},\mathscr{O}_X)$$

I remember seeing somewhere that this satisfies the graded commutativity (like the standard cup product on singular cohomology does). This means that if $s\in \check{H}^p$ and $t\in \check{H}^q$, $s\smile t=(-1)^{pq}t\smile s$.

I have tried to show this by fiddling with cocycles but I don't seem to be getting anywhere, but maybe I am just being silly.

So does graded commutativity hold? and if not why does it fail?

• Its true. Take a close look at the proof for singular cohomology. The combinatorics involved in the proof are identical to the Cech case. Every time you see an $n$-simplex, just replace by an $n$-fold intersection. – user113529 Apr 1 '16 at 5:17