Timeline for Is there a faithful functor from the freely generated bicartesian closed category to $\mathbf{Set}$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 2 at 17:41 | comment | added | varkor | @TimCampion: a bicartesian closed category is a cartesian closed category with finite coproducts. (The corresponding notion of functor is then clear.) | |
| Dec 12, 2023 at 10:02 | comment | added | Gro-Tsen | @QiaochuYuan This other question, that I just asked, is partly an attempt to make precise sense of the present one (which, as you point out, seems a bit strange). | |
| Oct 27, 2023 at 19:43 | comment | added | Qiaochu Yuan | Isn't $B$ just the category of finite sets? The objects that $B$ is required to contain are initial and terminal objects $0, 1$, then it has to be freely closed under finite products, coproducts, and exponentials. Since product distributes over coproduct, closure under finite products + coproducts produces exactly finite coproducts of copies of $1$, and the behavior of exponentials on two such objects is uniquely determined. Am I missing something? | |
| Oct 27, 2023 at 17:17 | comment | added | Tim Campion | What is a bicartesian closed category? What is a bicartesian closed functor? | |
| Oct 27, 2023 at 14:51 | comment | added | Alec Rhea | Unless ${\bf B}$ fails to be small, the functor $$X\longmapsto \{g:U\to X|g\in{\bf Hom}_{\bf B}\},$$ $$f:X\to Y\longmapsto f\circ,$$ yields a faithful functor into ${\bf Set}$. Further, even if this category fails to be small in e.g. $ZFC$, we can shift to a universe-based setting and get what we need by restricting attention to a version of ${\bf B}$ inside a fixed universe and be good to go. | |
| Oct 27, 2023 at 10:27 | history | asked | Johan Thiborg-Ericson | CC BY-SA 4.0 |