Timeline for Retract of a product
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2023 at 21:35 | comment | added | Achim Krause | Is it? For example, let $I=\mathbb{N} $ and $f$ the homomorphism with $f(x)=x_0+x_1$. Then both $J=\{0\}$ and $J=\{1\}$ seem contained, but not their intersection. | |
| Jun 11, 2023 at 22:01 | comment | added | YCor | @AchimKrause If $f:\mathbf{Z}^I\to \mathbf{Z}$ is a nonzero homomorphism, then $\{J: f|_{\mathbf{Z}^J}\neq 0\}$ is a $\sigma$-complete ultrafilter. | |
| Jun 11, 2023 at 13:34 | comment | added | Achim Krause | Interesting! How do you conversely turn a homomorphism $\mathbb{Z}^I\to\mathbb{Z}$ into an ultrafilter? | |
| Jun 11, 2023 at 7:12 | comment | added | YCor | If $\eta$ is a $\sigma$-complete ultrafilter on $I$, then $f\mapsto \lim_\eta f$ is a well-defined homomorphism $\mathbf{Z}^I\to\mathbf{Z}$. If $\eta$ is not principal, then this has "infinite support". | |
| Jun 11, 2023 at 7:10 | comment | added | YCor | @AchimKrause (To complement Jeremy's comment) if you precisely need that the canonical map $\mathbf{Z}^{(I)}\to\mathrm{Hom}(\mathbf{Z}^I,\mathbf{Z})$ is an isomorphism, then you precisely need the assumption that $I$ has cardinal $<$ than the smallest measurable cardinal (if it exists) — equivalently that every ultrafilter on $I$, $\sigma$-complete (=stable under countable intersections) is principal. This is true for $I$ countable, but also $I$ continuum, power of the continuum, and so on. | |
| Jun 9, 2023 at 15:39 | comment | added | Achim Krause | I just edited my answer: I addressed the question how the dualizing argument works, but unfortunately I also observed that the countable case does not formally imply the case of general $I$. So for now this is only an answer for countable $I$ (and maybe one can extend it somehow to the generality addressed by @JeremyRickard?) | |
| Jun 9, 2023 at 15:37 | history | edited | Achim Krause | CC BY-SA 4.0 |
added 1218 characters in body
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| Jun 9, 2023 at 15:08 | comment | added | user473423 | What do you mean by "dualize"? If you mean Potryagin Duality, I don't see how that helps. If your dualizing object is $\mathbb Z$, I see that you get that $Hom(G,\mathbb{Z})$ is a free abelian group. But how does that prove the original statement? | |
| Jun 9, 2023 at 14:55 | comment | added | Jeremy Rickard | For the statement in your first paragraph, I think you need $|I|$ to be smaller than the first measurable cardinal (if such a thing exists). | |
| Jun 9, 2023 at 14:18 | history | answered | Achim Krause | CC BY-SA 4.0 |