I'm trying to understand the footnote to Example 5.3 in [Wiegand - Sheaf cohomology of locally compact totally disconnected spaces](https://www.ams.org/journals/proc/1969-020-02/S0002-9939-1969-0253324-8/) which is about constructing a locally compact Hausdorff and totally disconnected space whose sheaf cohomology with constant sheaf $\mathbb Z/2\mathbb Z$ coefficients doesn't vanish. An explicit covering is given by compact open sets with no three intersecting and the sheaf cohomology is shown to coincide with the Čech cohomology over any covering by compact open sets. So things boil down to computing the Čech cohomology in this sheaf with respect to this covering.


In the end, since we are dealing essentially with boolean algebras, it turns into the following problem which I don't immediately see how to solve. If $Y$ is a set, let $B(Y)$ be the boolean algebra of finite and cofinite subsets of $Y$. Let $S$ be a countably infinite set and $T$ an uncountable set. Define a mapping $\Phi\colon B(T)^S\times B(S)^T\to 2^{S\times T}$ as follows. If $f\colon S\to B(T)$ and $g\colon T\to B(S)$ are maps, send $(f,g)$ to the subset of $S\times T$ consisting of all pairs $(s,t)$ with $s\in g(t)$ or $t\in f(s)$, but not both. Then I think the footnote is equivalent to the claim that $\Phi$ is not onto.


**Question**: why is $\Phi$ not onto?



A set theoretic question arising from trying to understand a sheaf cohomology question
This is not onto for any uncountably infinite $T$, even one much smaller than the power set (if the continuum hypothesis is false).


Fix $X \in 2^{ S \times T}$ such that the induced map $h \colon T \to 2^S$ where $h(t) = \{ s \mid (s,t) \in X\}$ (i.e. taking vertical fibers of $X$) has uncountable image. (For example we can take $T=2^S$ and choose $X$ to be $\{(s,t)| s\in T\}$ as LSpice suggested, so $h$ is the identity, or take $T$ to be a smaller uncountable subset of $2^S$ if one exists.)


Assume that $X$ does arise from the image of some $S$ and $T$ - we wil derive a contradiction from this. Let $U$ be the set of $s \in S$ such that $f(s)$ is cofinite (rather than finite). Then for each $s$, for all but finitely many $t$, we have $t \in f(s)$ if and only if $s\in U$. Since there are only countably many $s$, for all but countably many $t$, we have $t \in f(s)$ if and only if $s \in U$.


Thus, for all but countably many $t$, we have $s \in h(t)$ if and only if $s \in g(t)$ or $s\in U$, but not both. If $g(t)$ is finite then $h(t)$ is equal to $U$ up to finite error, and if $g(t)$ is cofinite then $h(t)$ is equal to the complement $U^c$ up to finite error.


In either case, there are countably many possibilities for $h(t)$, which together with the countably many $t$ excluded at first, gives countably many possibilities for $h(t)$ in total, contradicting our assumption.


