Timeline for Ultrafilter comonad on the category of Stone spaces
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 7, 2020 at 22:02 | vote | accept | Martin Brandenburg | ||
| Feb 7, 2020 at 21:59 | comment | added | Tim Campion | @MartinBrandenburg Exactly. This hold in general -- if $E \to A$ is any injection of sets, then $\beta E \to \beta A$ is also an injection, whose image consists of precisely those ultrafilters containing the image of $E$. And yes, the use of $p$ is crucial -- I think of $p$ as corresponding to a decomposition $B = \amalg_{a \in A} p^{-1}(a)$, and $f,g$ as choice functions selecting $f(a),g(a) \in p^{-1}(a)$ for each $a$. Then $E$ is the set where the choice functions agree, and $A \setminus E$ is the set where they disagree. | |
| Feb 7, 2020 at 21:57 | comment | added | Martin Brandenburg | Thanks! The proof for $g^{-1}(V)=\emptyset$ uses $p$. Why is it sufficient to prove $E \in \mathcal{F}$? Why need an ultrafilter $\mathcal{F}'$ on $E$ with $\mathcal{F}= i_* \mathcal{F}'$, where $i : E \to A$ is the inclusion. Do you take $\mathcal{F}' = \{T \cap E : T \in \mathcal{F}\}$? | |
| Feb 7, 2020 at 21:45 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
added 8 characters in body
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| Feb 7, 2020 at 19:39 | history | answered | Tim Campion | CC BY-SA 4.0 |