Timeline for Which categories are injective with respect to fully faithful functors?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2021 at 13:38 | answer | added | varkor | timeline score: 3 | |
| Apr 30, 2021 at 17:28 | answer | added | Tim Campion | timeline score: 2 | |
| Apr 29, 2021 at 20:46 | answer | added | Chris Schommer-Pries | timeline score: 11 | |
| Apr 29, 2021 at 14:58 | comment | added | მამუკა ჯიბლაძე | I believe there is weak (co)completeness that does not imply ordinary (co)completeness (requiring existence of initial/terminal cones without uniqueness). For example, I believe triangulated categories have all weak kernels and cokernels. I think any triangulated category with arbitrary (small) sums and products will be injective in the sense of (2). | |
| Apr 29, 2021 at 14:57 | comment | added | Chris Schommer-Pries | Let $\mathcal{A} = J$ be the free-walking isomorphism and $\mathcal{B} = pt$. Then $J \to pt$ is fully-faithful. So then to be injective in the strict sense requires that any isomorphism in $\mathcal{K}$ is an identity. I have heard that these are sometimes referred to as "gaunt" categories, and it is very restrictive. This seems very different from the pseudo case. | |
| Apr 29, 2021 at 13:47 | comment | added | Ivan Di Liberti | You are right Tim, I just wanted to pinpoint something in this spirit. | |
| Apr 29, 2021 at 13:45 | comment | added | Tim Campion | @IvanDiLiberti Thanks! I'm not requiring the functors $\mathcal A \to \mathcal K$ or $\mathcal B \to \mathcal K$ to be fully faithful, so the injectivity property here doesn't imply that $\mathcal K$ is "universal" in Trnkova's sense. But perhaps some of the ideas there are relevant. | |
| Apr 29, 2021 at 13:37 | comment | added | Ivan Di Liberti | I have no time today. Keywords: Trnková, Pultr, Universal Categories. | |
| Apr 29, 2021 at 13:29 | comment | added | Tim Campion | FWIW I think you can always choose the Kan extension in such a way to get a strict extension here as well. | |
| Apr 29, 2021 at 13:21 | comment | added | Tim Campion | @GregoryArone That's a good point, thanks! If $\mathcal K$ is complete, then the right Kan extension along a fully faithful $\mathcal A \to \mathcal B$ will restrict (up to isomorphism) to the original functor on $\mathcal A$. This relies on the fact that right Kan extensions in complete categories are pointwise -- it fails for general $\mathcal K$. This (+ the dual statement) means that that any complete or cocomplete category is pseudo-injective (and hence also lax / colax injective) with respect to fully faithful functors between small categories. | |
| Apr 29, 2021 at 13:17 | comment | added | Gregory Arone | Don't Kan extensions provide the kind of extension you are looking for, perhaps up to isomorphism, whenever $\mathcal K$ is either complete or cocomplete? | |
| Apr 29, 2021 at 13:08 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Apr 29, 2021 at 13:01 | history | asked | Tim Campion | CC BY-SA 4.0 |