Refinement choices differ on cochains but not on cohomology

$X$ is open in $X$ [F12], so the two members of $\mathcal{U}=(U_0,U_1)$ are open subsets with $U_0=U_1=X$ and $\bigcup \{U_0,U_1\}=X$, and the single member of $\mathcal{V}=(V_0)$ is an open subset with $V_0=X$; by the definition of an open cover [F7] both $\mathcal{U}$ and $\mathcal{V}$ are covers of $X$ by open subsets, and they are indexed by the linearly ordered sets $\{0<1\}$ and $\{0\}$ respectively, so their ordered Čech cochain complexes are defined [F5, F6]. The members of $\mathcal{U}$ are equal, which is permitted for an indexed cover: the condition is only that every $U_i$ be open and that the union of the family be $X$, and no injectivity of the index map is required.

The coarse cover $\mathcal{U}$ has $U_0\cap U_1=X$ and $U_0=U_1=X$, so $C^0(\mathcal{U},\mathcal{F})=\mathcal{F}(X)\oplus \mathcal{F}(X)$ and $C^1(\mathcal{U},\mathcal{F})=\mathcal{F}(X)$, while $C^p(\mathcal{U},\mathcal{F})=0$ for $p\ge 2$ because the two-element index set $\{0<1\}$ has no increasing triple [F5]. By the differential formula [F6] with $p=0$ and the increasing pair $(0,1)$, $({\delta }^0(a,b){)}_{01}=(-1{)}^{0}b{\mid }_{U_0\cap U_1}+(-1{)}^{1}a{\mid }_{U_0\cap U_1}=b-a,$ the restrictions being identities by [F10] since $U_0\cap U_1=X$; hence by [F4] ${\check{H}}^0(\mathcal{U},\mathcal{F})=\ker {\delta }^0=\{(a,a):a\in \mathcal{F}(X)\}\cong \mathcal{F}(X),$ the diagonal, and ${\check{H}}^1(\mathcal{U},\mathcal{F})=\mathcal{F}(X)\mathrm{/}\mathrm{im}{\delta }^0=\mathcal{F}(X)\mathrm{/}\mathcal{F}(X)=0$ because ${\delta }^0$ is surjective: ${\delta }^0(0,e)=e-0=e$ for every $e\in \mathcal{F}(X)$. In degrees $p\ge 2$ both $C^p$ and $C^{p+1}$ vanish, so ${\check{H}}^p(\mathcal{U},\mathcal{F})=0$ there as well [F4, F5].

The fine cover $\mathcal{V}$ has one member $V_0=X$, so $C^0(\mathcal{V},\mathcal{F})=\mathcal{F}(X)$ and $C^p(\mathcal{V},\mathcal{F})=0$ for every $p\ge 1$, the index set $\{0\}$ having no increasing pair and no longer tuple [F5]; the differentials of the complex of $\mathcal{V}$ are therefore the zero maps. By [F4], ${\check{H}}^0(\mathcal{V},\mathcal{F})=\ker ({\delta }^0:C^0\to C^1)=\mathcal{F}(X),{\check{H}}^p(\mathcal{V},\mathcal{F})=0(p\ge 1)\mathrm{.}$

Define $c,c^{\mathrm{\prime }}:\{0\}\to \{0,1\}$ by $c(0)=0$ and $c^{\mathrm{\prime }}(0)=1$. For $c$ the requirement of [F1] reads $V_0=X\subseteq U_{c(0)}=U_0=X$, and for $c^{\mathrm{\prime }}$ it reads $V_0=X\subseteq U_{c^{\mathrm{\prime }}(0)}=U_1=X$; both hold because $U_0=U_1=X$ [step 1.1]. Hence both $c$ and $c^{\mathrm{\prime }}$ are refinement functions from $\mathcal{V}$ to $\mathcal{U}$, and they are different functions since $c(0)=0\mathrm{\ne }1=c^{\mathrm{\prime }}(0)$.

Put $A:=\mathbb{Z}$ and let $\eta :A_{\mathrm{p}\mathrm{t}}\to A_X$ and $\theta :A_X\to {\underset{‾}{A}}_{\mathrm{l}\mathrm{o}\mathrm{c}}$ be the sheafification map and the identification with locally constant functions of [F8, F9]. Then $s_0:={\theta }_X({\eta }_X(0))$ and $s_1:={\theta }_X({\eta }_X(1))$ are elements of $\mathcal{F}(X)$, namely the constant functions with value $0$ and with value $1$ on $X$ [F8]. Since $X$ is nonempty, choose a point $x\in X$; the two functions take the different values $0\mathrm{\ne }1$ at $x$, so they are different functions and $s_0\mathrm{\ne }{s}_1$ in $\mathcal{F}(X)$; both ${\eta }_X$ and ${\theta }_X$ are group homomorphisms [F9], but only their values at $0$ and $1$ are needed here. Thus $(s_0,s_1)$ is a $0$-cochain of $\mathcal{U}$ with $s_0\mathrm{\ne }{s}_1$.

The group of $0$-cochains of $\mathcal{U}$ is $C^0(\mathcal{U},\mathcal{F})=\mathcal{F}(U_0)\times \mathcal{F}(U_1)=\mathcal{F}(X)\times \mathcal{F}(X)$ and that of $\mathcal{V}$ is $C^0(\mathcal{V},\mathcal{F})=\mathcal{F}(V_0)=\mathcal{F}(X)$, products over the single indices $0$ and $0,1$ respectively, in which the ordered and alternating models coincide [F5, F11]. Applying the formula of [F2] for the fine index $j_0=0$ to the cochain $(s_0,s_1)$ gives $(c^{\mathrm{♯}}(s_0,s_1){)}_0=s_{c(0)}{\mid }_{V_0}=s_0{\mid }_X=s_0,(c^{\mathrm{\prime }\mathrm{♯}}(s_0,s_1){)}_0=s_{c^{\mathrm{\prime }}(0)}{\mid }_{V_0}=s_1{\mid }_X=s_1,$ the restrictions being identities by [F10] because $V_0=X=U_0=U_1$. Since $s_0\mathrm{\ne }{s}_1$ [step 2.2], the two cochain maps differ in degree $0$: $c^{\mathrm{♯}}(s_0,s_1)=(s_0)\mathrm{\ne }(s_1)=c^{\mathrm{\prime }\mathrm{♯}}(s_0,s_1)$ as elements of $C^0(\mathcal{V},\mathcal{F})=\mathcal{F}(X)$.

The two cochain maps of [step 3.1] induce maps on cohomology in every degree by [F3]. In degree $0$ let $(a,a)\in {\check{H}}^0(\mathcal{U},\mathcal{F})$ be a diagonal class [step 1.2]; applying [F2] to the cochain $(a,a)$ gives $(c^{\mathrm{♯}}(a,a){)}_0=(a,a{)}_{c(0)}=a$ and $(c^{\mathrm{\prime }\mathrm{♯}}(a,a){)}_0=(a,a{)}_{c^{\mathrm{\prime }}(0)}=a$, since both components of the cochain are $a$; both values are cocycles of the complex of $\mathcal{V}$, whose degree one group is zero [step 1.3], so both induced maps send the class of $(a,a)$ to the class of $a$ under the isomorphism ${\check{H}}^0(\mathcal{U},\mathcal{F})\to {\check{H}}^0(\mathcal{V},\mathcal{F})=\mathcal{F}(X)$ which is $(a,a)\mapsto a$. In positive degrees ${\check{H}}^p(\mathcal{U},\mathcal{F})=0$ for $p\ge 1$ [step 1.2], so both induced maps are the zero map out of the zero group. Hence the induced maps on cohomology agree in every degree, as the refinement-homotopy theorem [F3] predicts, while the cochain maps themselves differ in degree $0$ [step 3.1].

Collected: $\mathcal{U}=(X,X)$ and $\mathcal{V}=(X)$ are open covers of the nonempty space $X$ [step 1.1]; the maps $c(0)=0$ and $c^{\mathrm{\prime }}(0)=1$ are both refinement functions from $\mathcal{V}$ to $\mathcal{U}$ [step 2.1]; the two induced cochain maps send the single $0$-cochain $(s_0,s_1)$ of [step 2.2] to the different elements $s_0$ and $s_1$ of $C^0(\mathcal{V},\mathcal{F})$ [step 3.1]; and yet the two induced maps on ${\check{H}}^p$ agree for every $p$, in degree $0$ as the isomorphism $(a,a)\mapsto a$ and in positive degrees as the zero map [step 4.1], in accordance with [F3]. Therefore the pair of covers alone does not determine a choice-independent map on Čech cochains: a chosen refinement function does determine a map, but changing $c$ can change that map, even though the induced map on cohomology is independent of it. No choice principle is used anywhere: the covers, the refinement functions and the sections $s_0,s_1$ are exhibited explicitly, the cochain maps are given by the restriction formula [F2], and the cohomology groups are computed from [F4] and [F5]. ∎