Three-open Čech sign cancellation

The increasing tuples of the linearly ordered set $\{0<1<2\}$ are the three singletons $(0),(1),(2)$ in degree $0$, the three pairs $(0,1),(0,2),(1,2)$ in degree $1$, the triple $(0,1,2)$ in degree $2$, and no increasing $(p+1)$-tuples for $p\ge 3$. Evaluating the product formula of [F1] on these tuples gives the four groups displayed in the statement, the product over an empty set of tuples being the zero group.

For $s=(s_0,s_1,s_2)\in C^0$ and the pair $(0,1)$ the formula of [F1] reads $({\delta }^{0}s{)}_{01}={\sum }_{j=0}^1(-1{)}^j{s}_{0\cdots \widehat{j}\cdots 1}=s_1{\mathrm{\mid }}_{U_{01}}-s_0{\mathrm{\mid }}_{U_{01}}$, and the same computation for the pairs $(0,2)$ and $(1,2)$ gives $({\delta }^{0}s{)}_{02}=s_2{\mathrm{\mid }}_{U_{02}}-s_0{\mathrm{\mid }}_{U_{02}}$ and $({\delta }^{0}s{)}_{12}=s_2{\mathrm{\mid }}_{U_{12}}-s_1{\mathrm{\mid }}_{U_{12}}$, that is, the three components of ${\delta }^0$ displayed in the statement with the signs $+,-$ attached to the second and first index respectively. For $c=(c_{01},c_{02},c_{12})\in C^1$ and the triple $(0,1,2)$ the formula reads $({\delta }^{1}c{)}_{012}={\sum }_{j=0}^2(-1{)}^j{c}_{0\cdots \widehat{j}\cdots 2}=c_{12}{\mathrm{\mid }}_{U_{012}}-c_{02}{\mathrm{\mid }}_{U_{012}}+c_{01}{\mathrm{\mid }}_{U_{012}}$, the alternating signs $+,-,+$ attaching to the omission of $j=0,1,2$. Since $C^3=0$ by [step 1.1], the differential ${\delta }^2$ and all higher differentials are the zero maps, since their targets are zero.

Composing the two computations of [step 2.1] gives $({\delta }^1({\delta }^{0}s){)}_{012}=(s_2-s_1){\mathrm{\mid }}_{U_{012}}-(s_2-s_0){\mathrm{\mid }}_{U_{012}}+(s_1-s_0){\mathrm{\mid }}_{U_{012}}$, all three terms lying in $\mathcal{F}(U_{012})$, the restrictions of the three components of ${\delta }^{0}s$ to the triple intersection. Expanding, the terms $s_2,s_1$ and $s_0$ each occur twice, once with each sign: $s_2-s_2=0$, $-s_1+s_1=0$ and $+s_0-s_0=0$ after collecting, so the sum is $0$ for every $0$-cochain $s$. This is the cancellation announced in the statement and the case $p=0$ of the general identity [F2].

The displayed groups of [step 1.1] and differentials of [step 2.1] are those of the ordered Čech complex of the three-open cover, and [step 3.1] verifies ${\delta }^1\circ {\delta }^0=0$ explicitly, in agreement with the general theorem [F2]; the vanishing of ${\delta }^2$ and of all higher differentials is [step 2.1], so all the remaining composites are zero as well and the cochain groups indeed form a complex. Applying the definition of fixed-cover cohomology [F3] in degree two, where ${\delta }^2=0$ and ${\delta }^1$ is displayed in [step 2.1], gives ${\check{H}}^2(\mathcal{U},\mathcal{F})=\ker {\delta }^2\mathrm{/}\mathrm{im}{\delta }^1=\mathcal{F}(U_{012})\mathrm{/}\mathrm{im}{\delta }^1$, and the groups in degrees zero and one are ${\check{H}}^0=\ker {\delta }^0$ and ${\check{H}}^1=\ker {\delta }^1\mathrm{/}\mathrm{im}{\delta }^0$. No choice principle is used: the tuples are finite and enumerated, the complex is finite and the cancellation is a finite computation in the abelian group $\mathcal{F}(U_{012})$. ∎