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I'm trying to understand the footnote to Example 5.3 in Wiegand - Sheaf cohomology of locally compact totally disconnected spaces 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?

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    $\begingroup$ @Dodd, if T is not uncountable then the map is onto because the sheaf cohomology vanishes so I suspect bad sets should have each s be related to and not related to uncountably many t's. $\endgroup$ – Benjamin Steinberg Jan 18 at 14:51
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    $\begingroup$ @Dodd, you can get the empty set by $(f,g)$ sending all elements to the empty set. You can get $S\times T$ by having $f$ send every one to the whole set and $g$ send everyone to the empty set. $\endgroup$ – Benjamin Steinberg Jan 18 at 14:54
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    $\begingroup$ In general, if you can get a set you can get its complement by pointwise complementing one of f or g. $\endgroup$ – Benjamin Steinberg Jan 18 at 14:56
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    $\begingroup$ Does it suffice to prove this for a specific pair of sets? If so, then the obvious thing seems to be to try to take $T = 2^S$ and to see if $\Phi$ hits the graph of $\in$. $\endgroup$ – LSpice Jan 18 at 14:59
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    $\begingroup$ @LSpice, that would be good enough and I thought of that relation but I don't see how to prove you can't get it. Maybe some diagonalization argument. $\endgroup$ – Benjamin Steinberg Jan 18 at 15:00
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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.

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  • $\begingroup$ The $t$ in the name of the induced map $X_t$ is just punctuation, right, not an element of $T$? That is, the map is $X_\text{punctuation}(t) = \{s \in S \mathrel: (s, t) \in X\}$? I think you also mean to speak of $X$ as the image of some $(f, g)$, rather than "of some $S$ and $T$". $\endgroup$ – LSpice Jan 18 at 15:24
  • $\begingroup$ I agree with Lspice. I first thought that $t$ was a fixed element of $t$ but now I see it is a function $\endgroup$ – Benjamin Steinberg Jan 18 at 15:30
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    $\begingroup$ Let me make sure I understood this. We view a subset of $S\times T$ as a function $X\colon T\to 2^S$. Since $T$ is uncountable and $2^S$ is too, we can construct an example of $X$ with image of $X$ uncountable. But if $X$ came from the image of $\Phi$, then the image of $X$ is countable. Is that correct? In particular, the example of @LSpice (which I had also considered) would work since this relation corresponds to the identity function $2^S\to 2^S$. $\endgroup$ – Benjamin Steinberg Jan 18 at 15:33
  • $\begingroup$ @Lspice Sorry, I am used to the notation $X_t$ for fibers and didn't realize that conflicts with calling it a function. $\endgroup$ – Will Sawin Jan 18 at 15:51
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    $\begingroup$ Right, and I think the argument could be stated more easily like this: Fix $\Phi$. Let $F$ be the set of $s$ with $f(s)$ finite, and let $C$ be the set of $s$ with $f(s)$ cofinite. Call $t\in T$ exceptional if there is some $s\in F$ s.t. $t\in f(s)$ or there is some $s\in C$ s.t. $t\notin f(s)$. There are countably many exceptional $t$, which contribute countably many sets to the image of $X$. For every non-exceptional $t$, $t\in f(s)$ for all $s\in C$ and $t\notin f(s)$ for $s\in F$. So $X(t)$ is the symmetric difference of $g(t)$ and $C$, and there are only countably many sets of this form. $\endgroup$ – Alex Kruckman Jan 18 at 15:52

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