Timeline for right adjoint for pullback along fibration
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 21, 2013 at 21:00 | comment | added | user2664 | For my part I read it, at least the statement is implicit, in Michael Shulman, Univalence For Inverse Diagrams p.10/11. | |
| Dec 21, 2013 at 15:33 | answer | added | Ronnie Brown | timeline score: 2 | |
| Dec 21, 2013 at 11:08 | comment | added | Michal R. Przybylek | (perhaps one may define $p^f(X)$ as a kind of a limit) | |
| Dec 21, 2013 at 11:07 | comment | added | Michal R. Przybylek | If "a fibration in a standard model structure on $\mathbf{Grpd}$" is a smart way of saying "fibration internal to $\mathbf{Grpd}$", then, I think, the fact easily follows from Conduche theorem ($p$ is exponentible in $\mathbf{Cat}$ iff $p$ is a Conduche fibration) --- since the pullback of a grupoid along a grupoid is a grupoid, the only thing to verify is that $p^f$ is grupoidal; but this is true since for any $h$, $\hom(h, p^f) \approx \hom(h \times_E f, p)$ is a grupoid. I am not aware of any simple fromula for $p^f(X)$ in case $f$ is not a fibration. | |
| Dec 20, 2013 at 21:00 | comment | added | David White | If you read it somewhere you should include that reference as part of the question. That way others would know where to look and what to reference when formulating an answer | |
| Dec 20, 2013 at 20:40 | comment | added | user2664 | I know this is true because I read it somewhere without proof and so I'm looking for a proof. | |
| Dec 20, 2013 at 20:34 | comment | added | Fernando Muro | It's kind of strange. If you know it's true, then you must know some proof, but you still ask about it, why is it so? | |
| Dec 20, 2013 at 18:07 | comment | added | user2664 | No the existence is not part of the question. It's true however as you may know $Grpd$ is not locally cartesian closed. But along fibration this right adjoint exists, this is the point. | |
| Dec 20, 2013 at 18:04 | comment | added | Fernando Muro | So is the existence of a right adjoint part of the question? If so, it would be nice if you could edit your question. | |
| Dec 20, 2013 at 18:03 | comment | added | user2664 | It seems it's true but I was not able to find a demonstration. Especially it would be nice to have an explicit and elementary construction of this right adjoint even if I would be also grateful for a theorem that solves the question. | |
| Dec 20, 2013 at 17:52 | comment | added | Fernando Muro | Do you happen to know that it has a right adjoint? | |
| Dec 20, 2013 at 17:48 | history | asked | user2664 | CC BY-SA 3.0 |