Čech cohomology of the two-arc cover of the circle

$p^{-1}(U_0)=(-1\mathrm{/}2,1\mathrm{/}2)+\mathbb{Z}$ and $p^{-1}(U_1)=(0,1)+\mathbb{Z}$ are unions of open intervals, hence open in $\mathbb{R}$, so $U_0$ and $U_1$ are open in $S^1$ [F7]. Both are nonempty and connected, being the images under the continuous map $p$ of the connected intervals $(-1\mathrm{/}2,1\mathrm{/}2)$ and $(0,1)$ [F8, F9]. Their union is all of $S^1$, because every real number differs by an integer from a point of $(-1\mathrm{/}2,1\mathrm{/}2)\cup (0,1)=(-1\mathrm{/}2,1)$. For the intersection, $p^{-1}(U_0\cap U_1)=((-1\mathrm{/}2,1\mathrm{/}2)+\mathbb{Z})\cap ((0,1)+\mathbb{Z})=((-1\mathrm{/}2,0)\cup (0,1\mathrm{/}2))+\mathbb{Z}$, so $U_0\cap U_1=p((-1\mathrm{/}2,0)\cup (0,1\mathrm{/}2))=p((1\mathrm{/}2,1))\cup p((0,1\mathrm{/}2))=C_2\cup C_1$, where $p(x+1)=p(x)$ was used [F6]. Each of $C_1,C_2$ is open, since $p^{-1}(C_1)=(0,1\mathrm{/}2)+\mathbb{Z}$ and $p^{-1}(C_2)=(1\mathrm{/}2,1)+\mathbb{Z}$ are open; each is nonempty and connected as the image of an interval [F8, F9]; and $C_1\cap C_2=\mathrm{\varnothing }$ because $(0,1\mathrm{/}2)+\mathbb{Z}$ and $(1\mathrm{/}2,1)+\mathbb{Z}$ are disjoint. So $(C_1,C_2)$ is a separation of $U_0\cap U_1$ in the sense of [F10], and every connected subset $T\subseteq U_0\cap U_1$ lies in $C_1$ or in $C_2$: if $T$ met both, then $T\cap C_1$ and $T\cap C_2$ would be nonempty disjoint relatively open subsets covering $T$, a separation of $T$ [F10]. Hence $C_1$ and $C_2$ are exactly the connected components of $U_0\cap U_1$.

By [F3] the sections of $\underset{‾}{\mathbb{Z}}$ over an open set are its locally constant $\mathbb{Z}$-valued functions. On each of the connected sets $U_0$, $U_1$, $C_1$, $C_2$ of [step 1.1] such a function is constant: its fibres are open, pairwise disjoint and cover the set, so if two distinct fibres were nonempty, one nonempty fibre and the union of all the other fibres would form a separation [F10]. Consequently the constant value of a section over $U_0$, over $U_1$, over $C_1$ and over $C_2$ is well defined, and the maps $f⟼f{\mathrm{\mid }}_{C_1}\text{’s value},f⟼f{\mathrm{\mid }}_{C_2}\text{’s value}$ give a bijection $\underset{‾}{\mathbb{Z}}(U_0\cap U_1)\to \mathbb{Z}\oplus \mathbb{Z}$: it is injective because a locally constant function on $U_0\cap U_1=C_1\cup C_2$ [step 1.1] is determined by its two constant values, and it is surjective because for $(m,n)\in \mathbb{Z}\oplus \mathbb{Z}$ the function equal to $m$ on $C_1$ and to $n$ on $C_2$ is locally constant, the two sets being open and disjoint [step 1.1]. The same argument with one connected set gives bijections $\underset{‾}{\mathbb{Z}}(U_0)\to \mathbb{Z}$ and $\underset{‾}{\mathbb{Z}}(U_1)\to \mathbb{Z}$ by constant value. All these bijections are group isomorphisms for the pointwise group structure of [F3], addition of locally constant functions being computed valuewise.

By [F1] applied to the two-member cover $\mathcal{U}=(U_0,U_1)$ the groups are $C^0(\mathcal{U},\underset{‾}{\mathbb{Z}})=\underset{‾}{\mathbb{Z}}(U_0)\oplus \underset{‾}{\mathbb{Z}}(U_1)$ and $C^1(\mathcal{U},\underset{‾}{\mathbb{Z}})=\underset{‾}{\mathbb{Z}}(U_0\cap U_1)$, and $C^p(\mathcal{U},\underset{‾}{\mathbb{Z}})=0$ for $p\ge 2$. Under the identifications of [step 2.1] the pair $(a,b)\in \mathbb{Z}\oplus \mathbb{Z}$ corresponds to the $0$-cochain whose components are the functions constantly equal to $a$ on $U_0$ and to $b$ on $U_1$; the difference $s_1{\mathrm{\mid }}_{U_0\cap U_1}-s_0{\mathrm{\mid }}_{U_0\cap U_1}$ of [F2] is then the function constantly equal to $b-a$ on $U_0\cap U_1$, hence restricts to the constant value $b-a$ on each of $C_1$ and $C_2$ and corresponds to $(b-a,b-a)\in \mathbb{Z}\oplus \mathbb{Z}$. Therefore ${\delta }^0(a,b)=(b-a,b-a)$ under the identifications, and ${\delta }^1=0$ since $C^2(\mathcal{U},\underset{‾}{\mathbb{Z}})=0$.

By [step 3.1] an element $(a,b)$ is a Čech $0$-cocycle exactly when $b-a=0$, that is exactly when $a=b$; hence the group of $0$-cocycles is the diagonal $\mathrm{\Delta }=\{(n,n):n\in \mathbb{Z}\}\cong \mathbb{Z}$ under $n\mapsto (n,n)$. The image of ${\delta }^0$ is $\{(b-a,b-a):a,b\in \mathbb{Z}\}=\mathrm{\Delta }$ as well. By [F4] applied to the complex of [step 3.1], whose terms $C^0,C^1$ are displayed there and whose terms in degrees $p\ge 2$ vanish, ${\check{H}}^0(\mathcal{U},\underset{‾}{\mathbb{Z}})=\ker {\delta }^0=\mathrm{\Delta }\cong \mathbb{Z},{\check{H}}^1(\mathcal{U},\underset{‾}{\mathbb{Z}})=(\mathbb{Z}\oplus \mathbb{Z})\mathrm{/}\mathrm{\Delta },{\check{H}}^p(\mathcal{U},\underset{‾}{\mathbb{Z}})=0(p\ge 2)\mathrm{.}$ The homomorphism $\phi :\mathbb{Z}\oplus \mathbb{Z}\to \mathbb{Z}$, $\phi (x,y)=x-y$, is surjective because $\phi (x,0)=x$, and $\ker \phi =\mathrm{\Delta }$; by the first isomorphism theorem [F11] it induces an isomorphism $(\mathbb{Z}\oplus \mathbb{Z})\mathrm{/}\mathrm{\Delta }\to \mathbb{Z}$, so ${\check{H}}^1(\mathcal{U},\underset{‾}{\mathbb{Z}})\cong \mathbb{Z}$.

Let $\kappa \in \underset{‾}{\mathbb{Z}}(U_0\cap U_1)$ be the locally constant function equal to $1$ on $C_1$ and to $0$ on $C_2$; it is a well-defined section by [step 2.1], and it corresponds to $(1,0)\in \mathbb{Z}\oplus \mathbb{Z}$ there. Every $1$-cochain is a cocycle because ${\delta }^1=0$ [step 3.1], so $\kappa$ has a class in ${\check{H}}^1(\mathcal{U},\underset{‾}{\mathbb{Z}})=(\mathbb{Z}\oplus \mathbb{Z})\mathrm{/}\mathrm{\Delta }$ of [step 4.1], namely the class of $(1,0)$; under the isomorphism induced by $\phi (x,y)=x-y$ this class corresponds to $\phi (1,0)=1$. Since $1$ generates $\mathbb{Z}$, the class $[\kappa ]$ generates ${\check{H}}^1(\mathcal{U},\underset{‾}{\mathbb{Z}})\cong \mathbb{Z}$.

By [F5] the restriction map $\mathrm{\Gamma }(S^1,\underset{‾}{\mathbb{Z}})\to {\check{H}}^0(\mathcal{U},\underset{‾}{\mathbb{Z}})$ is an isomorphism. The circle is connected: $S^1=p(\mathbb{R})$ is the image of the connected interval $\mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })$ [F8] under the continuous map $p$ [F9]. A locally constant $\mathbb{Z}$-valued function on the connected space $S^1$ is constant, by the argument of [step 2.1] applied to the whole circle: its fibres are open and partition $S^1$, so two distinct nonempty fibres would be a separation [F10]. Hence $\mathrm{\Gamma }(S^1,\underset{‾}{\mathbb{Z}})\cong \mathbb{Z}$ and ${\check{H}}^0(\mathcal{U},\underset{‾}{\mathbb{Z}})\cong \mathbb{Z}$, in agreement with the computation of [step 4.1]. This is the promised calculation for the two-arc cover: ${\check{H}}^0$ and ${\check{H}}^1$ are both isomorphic to $\mathbb{Z}$, with ${\check{H}}^1$ generated by the class of the overlap cocycle $\kappa$ of [step 5.1], and all higher Čech groups of the cover vanish. ∎