Timeline for The extension of the substitution map of the semigroup of variable words to its Stone–Čech compactification is a homomorphism
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2021 at 4:11 | vote | accept | andpe | ||
| Feb 11, 2021 at 19:41 | answer | added | Jordan Barrett | timeline score: 3 | |
| Feb 11, 2021 at 19:04 | comment | added | Benjamin Steinberg | (ctd) This extends by the universal property to $\lambda_s\colon \beta S\to \beta S$. Then for each $\sigma\in \beta S$ we have $\rho_{\sigma}\colon S\to \beta S$ by $\rho_{\sigma}(s) = \lambda_s(\sigma)$. This extends to a mapping $\rho_{\sigma}\colon \beta S\to \beta S$ and we can now define the product $\tau\cdot \sigma=\rho_{\sigma}(\tau)$. This should be easy to check is functorial. | |
| Feb 11, 2021 at 19:03 | comment | added | Benjamin Steinberg | As @JordanMitchellBarrett says, the $\beta$-semigroup is functorial. I think this is easier to see using the description of the product as limits. There are two ways to do the product based on whether you want left multiplication to be continuous or right multiplication. I'm not sure which convention corresponds to yours. I'll do once choice. For each $s\in S$, we have a continuous map $\lambda_s\colon S\to S$ by left translation $x\mapsto sx$. (Part 1) | |
| Feb 11, 2021 at 9:13 | comment | added | Jordan Barrett | Also, this feels like the kind of thing you should get "for free", in the following sense. I conjecture that for any semigroup homomorphism $f\colon S \to T$, its extension $\beta f\colon \beta S \to \beta T$ is a continuous homomorphism of topological semigroups. See if you can prove that, or look at Hindman and Strauss' book - I'm sure they prove it. | |
| Feb 11, 2021 at 8:45 | review | Suggested edits | |||
| Feb 11, 2021 at 10:17 | |||||
| Feb 11, 2021 at 8:42 | comment | added | Jordan Barrett | Then use the fact that ultrafilter quantifiers commute with all the logical connectives $\land$, $\lor$, $\to$, $\neg$. This makes proofs much easier - you don't have to deal with sets of sets of sets... | |
| Feb 11, 2021 at 8:41 | comment | added | Jordan Barrett | Do you know about ultrafilter quantifiers? You write $\mathcal{U}x\ \varphi(x)$ to mean that $\{ x: \varphi(x) \} \in \mathcal{U}$. Then, treat $\mathcal{U}x$ as a "quantifier" (like $\forall$ and $\exists$). The advantage is that $\mathcal{U}^\frown \mathcal{V}$ has a much simpler definition as $\{ A \subseteq S : \mathcal{U}x\ \mathcal{V} y\ (x^\frown y \in A) \}$. | |
| Feb 11, 2021 at 7:08 | history | asked | andpe | CC BY-SA 4.0 |