Map comparing Alexander-Spanier and Čech cohomology I'm trying to follow the argument in Bredon's Sheaf Theory (page 29) supposedly constructing an isomorphism between the Alexander-Spanier cohomology $H_{AS}^n (X; G)$ and Čech cohomology $\check H^n(X; G)$ for a space $X$ and an abelian group $G$ thought of as a constant sheaf. Specifically, Bredon constructs a map on the cochain groups that descends to cohomology. I'm having trouble seeing why this map is an isomorphism on cochains as claimed; in fact I believe I have a counterexample.
Specifically, $C_{AS}^n(X; G)$ is defined to be $A/A_0$, where $A$ is the group of all (possibly discontinuous) functions $X^{n+1} \to G$ and $A_0$ is the subgroup of functions that vanish on some neighborhood of the diagonal $X \hookrightarrow X^{n+1}$. The Čech cochain group $\check{C}^n(X; G)$ is defined to be the direct limit $\varinjlim \check C^n_{\mathfrak{U}}$ over all open covers $\mathfrak{U}$ of $X$ ordered by refinement, where given a cover $\mathfrak{U}$, 
$$\check{C}^n_{\mathfrak{U}} = \bigoplus_{U_0, \dots, U_n \in \mathfrak{U}} G(U_0 \cap \dots \cap U_n)$$
with the understanding that $G(\varnothing) = 0$. 
The map $C_{AS} \to \check{C}$ uses that a cofinal set of covers is given by covers of the form $\mathfrak{U} = \{U_x | x \in X\}$ indexed by $X$ with $U_x \ni x$, and refinements given by $\mathfrak{V} = \{V_x\}$ with $x \in V_x \subseteq U_x$. Then given an $f : X^{n+1} \to G$, we get $f_\infty  = \varinjlim f_{\mathfrak{U}} \in \check{C}^n(X; G)$ where 
$$f_{\mathfrak{U}}(U_{x_0} \cap \dots \cap U_{x_n}) = f(x_0, \dots, x_n) \; .$$
The book claims that the kernel of the map $f \mapsto f_\infty$ is supposed to be exactly $A_0$ (and has an associated claim about supports of $f$ and $f_\infty$ but I do not think this part is crucial).
I believe the following is a counterexample: Let $X = \{1/k | k \ge 1\} \cup \{0\}$, $G = \mathbb{Z}/2$ and 
$$f(x_0, \dots, x_n) = \begin{cases}0 & \text{if some } x_i = 0 \text{ or } x_0 = \dots = x_n\\ 1 & \text{otherwise.}\end{cases}$$ Then $f \notin A_0$ since any neighborhood of the diagonal contains $U^{n+1}$ for some neighborhood $U$ of $0$. But to compute $f_\infty$ we can take a further (cofinal) subset of covers $\mathfrak{U}$ where $U_{1/k} = \{1/k\}$ for each $k$. For such a $\mathfrak{U}$ the only way $U_{x_0} \cap \dots \cap U_{x_n}$ can be nonempty is if all the $x_i$'s are equal or at most one $x_i \ne 0$. So $f_{\mathfrak{U}} = 0$ and hence $f_\infty = 0$.
Am I misinterpreting something? Or does this statement need more assumptions on $X$ (or $G$)? Note that the counterexample has $X$ a compact subset of $\mathbb{R}$, although not (locally) connected. Is it still true that the cohomology is isomorphic even if the chain groups are not?
 A: I read through Bredon’s book, and at page 29, when dealing with Cech cohomology for paracompact spaces, he restricts to locally finite open coverings. I think it is the open covering you gave cause the issue, it is not locally finite at 0. In fact, if you use a finite covering of X, then the image $f_\infty$ is nontrivial, since for k big enough, its neighbor coincide with the neighborhood of 0, thus Uk intersect Ul non empty, $f_\infty{(1/k,1/l)}=f(1/k,1/l)=1$. 
A: $\!$Hi Ronno, just found this old question, small world. A few comments:

*

*I agree that Bredon's book is mistaken here.

*Define a rigid open cover to be a collection of open sets $U_x \ni x$ for all $x \in X$, with the additional property that the set $\{U_x | x \in X\}$ is a locally finite cover of $X$. If $X$ is paracompact then rigid open covers are cofinal among all open covers, so rigid open covers can be used to compute Cech cohomology. As you pointed out in a comment to Alan Zhang, rigid open covers will not in general be cofinal among all open covers indexed by $X$, but this is not quite the property we need, anyway. It seems to me that Bredon's argument will work to prove an isomorphism between Cech and Alexander-Spanier cohomology on paracompact spaces, once we make the choice of computing Cech cohomology using rigid open covers.

*The above is a bit unsatisfying since paracompactness shouldn't be needed to compare Cech and Alexander-Spanier cohomology. There's a surprising theorem of Dowker ("Homology groups of relations", Annals 1952) that says that Cech cohomology and Alexander-Spanier cohomology are isomorphic on any topological space whatsoever. Usually one expects at least some kind of point-set assumption for "reasonable" cohomology theories to coincide! Both Cech and Alexander-Spanier theories are "badly behaved" for spaces that are not paracompact Hausdorff, but apparently they are badly behaved in exactly the same way. I haven't tried digging into Dowker's argument, though. (Edited to add: Actually Dowker's theorem is quite pretty and it seems much more like the "right" approach than what Bredon does.)

