Timeline for Images of complemented subobjects in toposes
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 23, 2021 at 22:30 | vote | accept | Mendieta | ||
| Jun 23, 2021 at 19:35 | comment | added | Mendieta | We could assume furthermore that $f$ is hyperconnected. Then the Heyting algebra of subobjects of ${f^* \mathbb{N}}$ would be Boolean. | |
| Jun 23, 2021 at 19:31 | comment | added | მამუკა ჯიბლაძე | @Mendieta oops sorry again. Must be more careful. | |
| Jun 23, 2021 at 19:30 | comment | added | Mendieta | @მამუკაჯიბლაძე, the subobject $u$ is complemented by hypothesis. | |
| Jun 23, 2021 at 19:21 | comment | added | მამუკა ჯიბლაძე | @Mendieta Sorry you are right. What I said does not matter. What matters is that one can modify the answer to include any non-Boolean topos $E$ that admits a geometric morphism to a Boolean topos $S$. Take $u$ any non-complemented subobject of some object $x$ in $E$, and the rest as in the answer: take $s$ the terminal object of $S$ and $\pi$ the identity morphism of $x$. | |
| Jun 23, 2021 at 17:29 | comment | added | Mendieta | @მამუკაჯიბლაძე What you say is true, but the question still makes sense because E need not be Boolean. | |
| Jun 23, 2021 at 6:46 | comment | added | მამუკა ჯიბლაძე | @Mendieta If $S$ is Boolean (which is weaker than having AC), every subobject of every object is complemented. | |
| Jun 22, 2021 at 23:43 | comment | added | Mendieta | Thanks very much for your reply, @Andreas Blass. I'd be happy to mark my question as answered; but I think it would be more interesting to edit it. What if we assume that $S$ is Boolean or satisfies Choice? (Shouldn't $\pi$ behave as a codiagonal?) | |
| Jun 22, 2021 at 23:17 | history | answered | Andreas Blass | CC BY-SA 4.0 |