The Čech differential squares to zero

First take $p\ge 0$. Fix an increasing $(p+3)$-tuple $i_0<\cdots <i_{p+2}$, which indexes a component of $C^{p+2}$, the target of ${\delta }^{p+1}{\delta }^p$. If no such tuple exists, then $C^{p+2}=0$ and the identity is vacuous. If the full intersection is empty, its section group is zero by [F3]. Otherwise the component $({\delta }^{p+1}{\delta }^{p}s{)}_{i_0\cdots i_{p+2}}$ is computed in the single abelian group $\mathcal{F}(U_{i_0}\cap \cdots \cap U_{i_{p+2}})$.

Applying [F1] twice gives $({\delta }^{p+1}{\delta }^{p}s{)}_{i_0\cdots i_{p+2}}={\sum }_{j=0}^{p+2}(-1{)}^j({\delta }^{p}s{)}_{i_0\cdots \widehat{i_j}\cdots i_{p+2}}{\mid }_{U_{i_0}\cap \cdots \cap U_{i_{p+2}}}\mathrm{.}$ Expanding each inner differential produces terms indexed by ordered pairs of distinct positions $j,k\in \{0,\dots ,p+2\}$; each is the restriction of the $p$-cochain component $s_{i_0\cdots \widehat{i_j}\cdots \widehat{i_k}\cdots i_{p+2}}$, whose tuple has $p+1$ indices. The coefficient for omitting $i_j$ first and then $i_k$ is $(-1{)}^{j+k-1}$ when $j<k$, and $(-1{)}^{j+k}$ when $j>k$. Compatibility of restrictions [F2] places all terms in the section group of the full intersection.

For $j<k$, pair the term omitting $i_j$ then $i_k$ with the term omitting $i_k$ then $i_j$. By [F2] these are restrictions of the same $p$-cochain component to the full intersection, while their coefficients $(-1{)}^{j+k-1}$ and $(-1{)}^{j+k}$ sum to zero. Every ordered pair occurs in exactly one such pair. Thus every component of ${\delta }^{p+1}{\delta }^{p}s$ vanishes.

The tuple was arbitrary, so step 2.1 proves the identity for $p\ge 0$. For $p<0$, the convention $C^p=0$ and ${\delta }^p=0$ makes the composite zero. Hence ${\delta }^{p+1}\circ {\delta }^p=0$ in every degree, and the Čech cochains form a cochain complex.