Ordered and alternating Čech complexes agree

Fix $p\ge 0$ and a tuple $K=(i_0,\dots ,i_p)$ of pairwise distinct indices. There is exactly one permutation ${\tau }_K\in S_{p+1}$ with $i_{{\tau }_K(0)}<\cdots <i_{{\tau }_K(p)}$, because the indices are pairwise distinct; call it the sorting permutation of $K$ and write $\mathrm{sort}(K):={\tau }_K\cdot K$ for the increasing rearrangement. For $\sigma \in S_{p+1}$ one has ${\tau }_{\sigma K}={\sigma }^{-1}{\tau }_K$: the entry of $\sigma \cdot K$ at position ${\tau }_{\sigma K}(j)$ is $i_{\sigma ({\tau }_{\sigma K}(j))}$, and this is increasing in $j$ exactly when $\sigma {\tau }_{\sigma K}={\tau }_K$; taking signs gives $\mathrm{sgn}({\tau }_{\sigma K})=\mathrm{sgn}({\sigma }^{-1})\mathrm{sgn}({\tau }_K)=\mathrm{sgn}(\sigma )\mathrm{sgn}({\tau }_K)$ by multiplicativity of the sign [F5] and [F4]. Also $\mathrm{sort}(\sigma K)=\mathrm{sort}(K)$, since $K$ and $\sigma \cdot K$ have the same underlying set of indices. Now define $e^p:C^p(\mathcal{U},\mathcal{F})\to {\prod }_{K\in I^{p+1}}\mathcal{F}(\bigcap K)$ by $(e^{p}t{)}_K:=\mathrm{sgn}({\tau }_K)t_{\mathrm{sort}(K)}$ for pairwise distinct $K$ and $(e^{p}t{)}_K:=0$ otherwise; this is well defined because $\bigcap \mathrm{sort}(K)=\bigcap K$ as open sets. The map $e^p$ is a homomorphism of abelian groups, it vanishes on tuples with a repeated index by construction, and for $\sigma \in S_{p+1}$ one has $(e^{p}t{)}_{\sigma K}=\mathrm{sgn}({\tau }_{\sigma K})t_{\mathrm{sort}(\sigma K)}=\mathrm{sgn}(\sigma )\mathrm{sgn}({\tau }_K)t_{\mathrm{sort}(K)}=\mathrm{sgn}(\sigma )(e^{p}t{)}_K$, so $e^p$ takes values in ${\tilde{C}}^p(\mathcal{U},\mathcal{F})$; moreover ${\rho }^p{e}^p=\mathrm{id}$, because ${\tau }_K$ is the identity for increasing $K$.

Fix $\sigma \in S_{p+2}$, a tuple $K=(i_0,\dots ,i_{p+1})$ of pairwise distinct indices and a position $a\in \{0,\dots ,p+1\}$. Let ${\iota }_a$ be the order-preserving bijection from the positions of the tuple $K\setminus a:=(i_0,\dots ,\widehat{i_a},\dots ,i_{p+1})$ onto the set $\{0,\dots ,p+1\}\setminus \{a\}$, and define the permutation ${\sigma }_a\in S_{p+1}$ of the positions of $K\setminus a$ by ${\sigma }_a:={\iota }_{\sigma (a)}^{-1}\circ \sigma \circ {\iota }_a$. Claim: $\mathrm{sgn}({\sigma }_a)=\mathrm{sgn}(\sigma )(-1{)}^{a+\sigma (a)}$. Since ${\iota }_a$ and ${\iota }_{\sigma (a)}$ preserve order, $\mathrm{inv}({\sigma }_a)$ counts the pairs of positions $c<c^{\mathrm{\prime }}$ of $K$ with $c,c^{\mathrm{\prime }}\mathrm{\ne }a$ and $\sigma (c)>\sigma (c^{\mathrm{\prime }})$, that is $\mathrm{inv}({\sigma }_a)=\mathrm{inv}(\sigma )-A-B$ with $A:=\mathrm{\#}\{c<a:\sigma (c)>\sigma (a)\}$ and $B:=\mathrm{\#}\{c^{\mathrm{\prime }}>a:\sigma (a)>\sigma (c^{\mathrm{\prime }})\}$. Among the $a$ positions $c<a$ exactly $A$ satisfy $\sigma (c)>\sigma (a)$ and the remaining $a-A$ satisfy $\sigma (c)<\sigma (a)$; of the $\sigma (a)$ values below $\sigma (a)$ exactly those $a-A$ are images of positions $<a$, so the remaining $\sigma (a)-(a-A)$ of them are images of positions $>a$, that is $B=\sigma (a)-a+A$. Hence $\mathrm{inv}({\sigma }_a)=\mathrm{inv}(\sigma )+a-\sigma (a)-2A$, and reducing modulo $2$ and using $\mathrm{sgn}=(-1{)}^{\mathrm{inv}}$ [F4] gives the claim.

For an increasing tuple $i_0<\cdots <i_{p+1}$ every tuple $i_0\cdots \widehat{i_j}\cdots i_{p+1}$ obtained by deleting one entry is increasing again, so restricting to increasing tuples changes none of the values: expanding both sides with the differential formula [F1] gives $({\rho }^{p+1}{\delta }^{p}s{)}_{i_0\cdots i_{p+1}}={\sum }_{j=0}^{p+1}(-1{)}^j{s}_{i_0\cdots \widehat{i_j}\cdots i_{p+1}}{\mathrm{\mid }}_{U_{i_0}\cap \cdots \cap U_{i_{p+1}}}=({\delta }^p{\rho }^{p}s{)}_{i_0\cdots i_{p+1}}$ term by term, with the same restriction maps on both sides [F3].

Conversely $e^p{\rho }^p=\mathrm{id}$ on ${\tilde{C}}^p(\mathcal{U},\mathcal{F})$. Let $s$ be alternating and let $K$ be a tuple of pairwise distinct indices; applying the second alternating condition to the sorting permutation ${\tau }_K$ gives $s_{\mathrm{sort}(K)}=s_{{\tau }_K\cdot K}=\mathrm{sgn}({\tau }_K)s_K$, hence $s_K=\mathrm{sgn}({\tau }_K)s_{\mathrm{sort}(K)}=(e^p({\rho }^{p}s){)}_K$, because $\mathrm{sgn}({\tau }_K)$ is its own inverse [F4] and ${\rho }^{p}s$ records the values of $s$ on increasing tuples; on a tuple with a repeated index both $s_K$ and $(e^p({\rho }^{p}s){)}_K$ are $0$. Together with ${\rho }^p{e}^p=\mathrm{id}$ from [step 1.1] this shows that ${\rho }^p$ is a bijection with inverse $e^p$, hence an isomorphism of abelian groups, and proves assertion 1 of the statement. [F4, step 1.1]

Let $s\in {\tilde{C}}^p(\mathcal{U},\mathcal{F})$, let $K$ have pairwise distinct entries and length $p+2$, and let $\sigma \in S_{p+2}$. Deleting position $a$ in $\sigma K$ deletes the entry $i_{\sigma (a)}$, so the definition in step 1.2 gives $(\sigma K)\setminus a={\sigma }_a\cdot (K\setminus \sigma (a))$. Thus $s_{(\sigma K)\setminus a}=\mathrm{sgn}({\sigma }_a)s_{K\setminus \sigma (a)}$. Including all restrictions to the common full intersection, the sign identity in step 1.2 gives $({\delta }^{p}s{)}_{\sigma K}={\sum }_a(-1{)}^a\mathrm{sgn}({\sigma }_a)s_{K\setminus \sigma (a)}=\mathrm{sgn}(\sigma ){\sum }_a(-1{)}^{\sigma (a)}{s}_{K\setminus \sigma (a)}=\mathrm{sgn}(\sigma )({\delta }^{p}s{)}_K\mathrm{.}$ The last sum is reindexed by $b=\sigma (a)$. Restrictions commute with these signs by [F3], so this proves alternation on distinct tuples.

Next, ${\delta }^{p}s$ vanishes on a tuple $K$ with a repeated entry. Choose positions $u<v$ with $i_u=i_v$. In the sum of the statement every term with $j\mathrm{\notin }\{u,v\}$ contributes $s_{K\setminus j}=0$, because the equal pair survives in $K\setminus j$; and if $K\setminus u$ has a repeated entry then $K\setminus v$ has the same multiset of entries and also has one, so the two remaining terms vanish as well. Otherwise $K\setminus u$ and $K\setminus v$ have pairwise distinct entries and equal multisets, and $K\setminus u$ arises from $K\setminus v$ by moving the entry $i_v=i_u$ from position $u$ to position $v-1$ past the entries $i_{u+1},\dots ,i_{v-1}$, that is by the $(v-u)$-cycle of positions $(uu+1\cdots v-1)$, whose sign is $(-1{)}^{v-u-1}$ [F6]; hence $s_{K\setminus u}=\mathrm{sgn}(\pi )s_{K\setminus v}$ for that cycle $\pi$ and $(-1{)}^u{s}_{K\setminus u}+(-1{)}^v{s}_{K\setminus v}=[(-1{)}^u(-1{)}^{v-u-1}+(-1{)}^v]s_{K\setminus v}=[(-1{)}^{v-1}+(-1{)}^v]s_{K\setminus v}=0$. In every case $({\delta }^{p}s{)}_K=0$, all values being taken in the group $\mathcal{F}(U_{i_0}\cap \cdots \cap U_{i_{p+1}})$ [F3].

Combining: the formula of the statement is additive in $s$ and, by [step 2.2] and [step 3.1], it maps ${\tilde{C}}^p(\mathcal{U},\mathcal{F})$ into ${\tilde{C}}^{p+1}(\mathcal{U},\mathcal{F})$; by [step 1.3] it satisfies ${\rho }^{p+1}{\delta }^p={\delta }^p{\rho }^p$; and by [step 1.1] and [step 2.1] the map ${\rho }^p$ is bijective with inverse $e^p$ in every degree. Hence ${\rho }^{\bullet }$ is an isomorphism of cochain complexes and assertion 2 holds. Finally, for $s\in {\tilde{C}}^p(\mathcal{U},\mathcal{F})$ one has ${\rho }^{p+2}({\delta }^{p+1}{\delta }^{p}s)=({\delta }^{p+1}{\delta }^p)({\rho }^{p}s)=0$ by [F2], and ${\rho }^{p+2}$ is injective by [step 2.1], so ${\delta }^{p+1}{\delta }^p=0$; in the negative degrees both complexes are $0$ by their conventions, and the extension formula exhibits the values of any cochain at arbitrary tuples of $U$-indices. ∎