Timeline for Generators in the sense of Freyd and Kelly
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2014 at 19:08 | vote | accept | fosco | ||
| Dec 29, 2013 at 14:51 | comment | added | fosco | Ah! I understood. FK define a precise notion of a family "being in $\mathfrak E$", which is precisely the one you are stating. Sorry, this question was boring and ill-posed. | |
| Dec 29, 2013 at 14:45 | comment | added | fosco | The problem is precisely this, I misunderstood something as I'm left with something which is blatantly false: but how could one interpret the sentence "the family of all morphisms $G\to A$ with domain $G\in\cal G$ is in $\mathfrak E$" in a different manner? The "family" taken as a single set? But it is not an object of $\cal A$, it doesn't make any sense. | |
| Dec 29, 2013 at 14:39 | comment | added | Zhen Lin | You have misunderstood something: we are not claiming that each individual morphism is in $\mathcal{E}$, but that they are "jointly" in $\mathcal{E}$. Your "proof" is also mistaken. | |
| Dec 29, 2013 at 14:37 | comment | added | fosco | Please notice that I edited the OP. Thank you for your time! | |
| Dec 29, 2013 at 14:30 | comment | added | fosco | Sure, this is exactly the remark at the end of page 177 in FK paper. But my problem is different: FK say that a generator for a category is the gadget I defined before, and under reasonable hypotheses a generator/generating subcat is the same as a separator/separating subcat. But this is false! The point is a separator in Set, and nevertheless it is not true that any arrow $G=*\to A$ is an epi. On the other hand it is true that the arrow $k_A\colon \coprod_{G\to A}G\to A$ is an epi; where am I lost? | |
| Dec 29, 2013 at 14:23 | history | answered | Zhen Lin | CC BY-SA 3.0 |