Fixed-cover Čech can miss derived cohomology

Let $X$ be a topological space, let $\mathcal{F}$ be a sheaf of abelian groups on $X$ and let $\mathcal{U}=(U_i{)}_{i\in I}$ be an open cover of $X$ indexed by a linearly ordered set, with fixed-cover Čech cohomology ${\check{H}}^{\bullet }(\mathcal{U},\mathcal{F})$ (Fixed-cover Čech cohomology). The cover is part of the data: the groups ${\check{H}}^p(\mathcal{U},\mathcal{F})$ are the cohomology of the ordered Čech cochain complex of $\mathcal{U}$ alone (Ordered Čech cochain complex of a cover), and the definition refers to no other cover of $X$.

1. The one-member cover has no positive cochains. Take $\mathcal{U}$ to be the cover $\{X\}$, that is $I=\{0\}$ and $U_0=X$. The index set has the single increasing $0$-tuple $(0)$ and no increasing $(p+1)$-tuple for $p\ge 1$, and a product over an empty index set is the trivial group, so $C^0(\mathcal{U},\mathcal{F})=\mathcal{F}(X),C^p(\mathcal{U},\mathcal{F})=0(p\ge 1)\mathrm{.}$ Hence ${\delta }^0$ maps into the zero group and all higher differentials vanish, giving ${\check{H}}^0(\mathcal{U},\mathcal{F})=\mathrm{\Gamma }(X,\mathcal{F}),{\check{H}}^p(\mathcal{U},\mathcal{F})=0(p\ge 1),$ for every abelian sheaf $\mathcal{F}$ on every topological space $X$.

2. The comparison need not be an isomorphism. Under the Axiom of Choice (The Axiom of Choice) the comparison map ${\phi }_{\mathcal{U}}^p:{\check{H}}^p(\mathcal{U},\mathcal{F})\to H^p(X,\mathcal{F})$ is defined for every cover (Canonical map from fixed-cover Čech to sheaf cohomology), and it is an isomorphism in every degree when $\mathcal{U}$ is $\mathcal{F}$-acyclic (Leray acyclic-cover comparison). The acyclicity of the cover on its nonempty finite intersections cannot be dropped. On the circle $X=S^1$ with $\mathcal{F}$ the constant sheaf $\mathbb{Z}$ the one-member cover has ${\check{H}}^1=0$ by part 1 while $H^1(S^1,\mathbb{Z})\mathrm{\ne }0$, so ${\phi }_{\mathcal{U}}^1$ is the zero homomorphism out of the zero group into a nonzero group. That cover is not $\mathbb{Z}$-acyclic, its only nonempty finite intersection being $S^1$ itself; the companion examples page of this pair computes this cover and the two-arc cover of the circle explicitly and records a nonzero class in $H^1(S^1,\mathbb{Z})$.

3. Where the nonzero classes come from. Nonzero first cohomology is not an exotic phenomenon. For a short exact sequence $0\to {\mathcal{F}}^{\mathrm{\prime }}\to \mathcal{F}\to {\mathcal{F}}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ of abelian sheaves the long exact sequence (Long exact sequence of sheaf cohomology) begins $H^0(X,\mathcal{F})⟶H^0(X,{\mathcal{F}}^{\mathrm{\prime }\mathrm{\prime }})\stackrel{{\mathrm{\partial }}^0}{\to }{H}^1(X,{\mathcal{F}}^{\mathrm{\prime }}),$ and $H^0(X,-)$ is canonically the global-sections functor (Degree-zero sheaf cohomology is global sections). Global sections are left exact, but an epimorphism of sheaves need not be surjective on global sections (Global sections are left exact but need not preserve epimorphisms); whenever that surjectivity fails, the connecting homomorphism has nonzero image and $H^1(X,{\mathcal{F}}^{\mathrm{\prime }})\mathrm{\ne }0$. For such a sheaf ${\mathcal{F}}^{\mathrm{\prime }}$ the one-member cover of $X$ already exhibits the failure of part 2 in degree one, its source ${\check{H}}^1$ being the zero group.

4. Conclusion. The fixed-cover groups are therefore not a function of the pair $(X,\mathcal{F})$ alone: on the circle with the constant sheaf $\mathbb{Z}$ the one-member cover gives ${\check{H}}^1=0$, which disagrees with $H^1(S^1,\mathbb{Z})\mathrm{\ne }0$, while the two-arc cover of the circle gives ${\check{H}}^1\cong \mathbb{Z}$ (companion examples page). The equality ${\check{H}}^p(\mathcal{U},\mathcal{F})\cong H^p(X,\mathcal{F})$ is a theorem about the cover, available under the acyclicity hypothesis on its finite intersections (Leray acyclic-cover comparison), and not a formal consequence of the definition of ${\check{H}}^p(\mathcal{U},\mathcal{F})$ alone.