Timeline for A dual notion to Lawvere-Tierney operators for geometric surjections?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2014 at 8:26 | comment | added | მამუკა ჯიბლაძე | As for quotient geometric morphisms - there is some work done; they are related to exponentiable ones (Johnstone & Joyal, Niefield, Plewe). There are some relatively well understood subclasses, e. g. the so called triquotient maps. | |
| Nov 9, 2014 at 8:25 | comment | added | მამუკა ჯიბლაძე | Btw although there is no quotient classifier, for each object $X$ one may construct the object of its quotients (it is isomorphic to the object of equivalence relations on $X$) | |
| Nov 9, 2014 at 8:24 | comment | added | მამუკა ჯიბლაძე | Well I don't want to go that far. It might be possible to do something, and it would be very interesting to do it. I am just trying to explain why is it essentially more complicated. | |
| Nov 9, 2014 at 8:24 | comment | added | Ali Lahijani | About your last sentence, that surjections are not necessarily quotients, are there a subset of geometric surjections that do just that? An intermediate step before full surjections and hopefully a three-way factorization system? | |
| Nov 9, 2014 at 8:19 | comment | added | Ali Lahijani | So, you are saying, since there is no device like a subobject classfier that classifies quotients, we cannot hope for an analogous result. | |
| Nov 9, 2014 at 7:56 | comment | added | მამუკა ჯიბლაძე | Another complication stems from the fact that while subtoposes of $\mathcal E$ are defined over $\mathcal E$, quotient toposes of $\mathcal E$ are defined "under $\mathcal E$", in a sense $\mathcal E$ "does not see them". Another related thing is that surjections are not necessarily quotients, so passing to the surjective image you not only "make $\mathcal E$ smaller" but also endow it with additional structure. | |
| Nov 9, 2014 at 7:52 | comment | added | მამუკა ჯიბლაძე | I see. Well, by analogy with the "toy case" this must be essentially more difficult. Subsets of a set form a Boolean algebra, while the lattice of quotient sets is much more nasty. It is also very special in a sense, but not even modular (in most cases). | |
| Nov 9, 2014 at 7:47 | history | edited | Ali Lahijani | CC BY-SA 3.0 |
added 20 characters in body
|
| Nov 9, 2014 at 7:41 | comment | added | Ali Lahijani | You're right. I should be asking about surjections out of $\mathcal{E}$. Even an ionad is a surjection out of $Set^X$. | |
| Nov 9, 2014 at 6:00 | comment | added | მამუკა ჯიბლაძე | Do you really mean surjections into $\mathcal E$? Or surjections out of $\mathcal E$? | |
| Nov 9, 2014 at 2:26 | history | asked | Ali Lahijani | CC BY-SA 3.0 |