Timeline for Is a composite of (co)monadic adjunctions (co)monadic?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 11 at 9:38 | comment | added | Todd Trimble | @KevinCarlson Oh, this is a nice observation -- I'll remember this. Thanks! | |
| Jun 10 at 23:05 | comment | added | Kevin Carlson | @ToddTrimble This seemed like a reasonable conversation in which to record a confusion of my own I've (hopefully) straightened out: plain graphs are monadic over $\mathbf{Set}$ by taking the product of vertices and edges (rather than their sum or just the edges), and the analogous fact is true for the category of $C$-sets on any category with finitely many objects, because both $[C,\mathbf{Set}]\to [\mathrm{ob} C,\mathbf{Set}]$ and $\prod:[\mathrm{ob} C,\mathbf{Set}]\to \mathbf{Set}$ are crudely monadic when $\mathrm{ob} C$ is finite. | |
| Dec 20, 2023 at 2:38 | comment | added | Todd Trimble | @DanielSchäppi Yeah, TTT certainly misspoke there, unless by "graph" Barr and Wells meant reflexive graph (which I doubt they did). In particular, as you seem to recognize, the arrow functor to sets can't be monadic because it's not faithful (since it cannot detect behavior at isolated points of a graph). | |
| Mar 6, 2016 at 16:54 | comment | added | Todd Trimble | Right, the usual example is that $\mathbf{Cat}$ is monadic over reflexive graphs, and reflexive graphs are monadic over $\mathbf{Set}$ by sending a reflexive graph to its set of arrows; indeed the category of reflexive graphs is equivalent to the category of actions of the monoid with two generators $s, t$ subject to the relations $s s = s = t s$ and $s t = t = t t$. | |
| Jan 7, 2014 at 18:23 | vote | accept | Akhil Mathew | ||
| Jan 7, 2014 at 17:14 | comment | added | Daniel Schäppi | Sorry, the functor in question sends a graph to the disjoint union of its set of arrows and its set of vertices, not just its set of arrows. The two nontrivial elements $s,t$ of the monoid act trivially on the set of objects, and send arrows to their source and target respectively. The usual relations that source of the target is target etc. have to be imposed. | |
| Jan 7, 2014 at 16:58 | comment | added | Daniel Schäppi | Well, it is also monadic over $\mathbf{Set}\times \mathbf{Set}$, but the functor which sends a graph to its set of arrows is also monadic. In fact, there is a three element monoid in $\mathbf{Set}$ whose category of actions is equivalent to graphs (see exercise (GRMN) on page 107 of "TTT"). | |
| Jan 7, 2014 at 16:41 | comment | added | Zhen Lin | Are graphs really monadic over $\mathbf{Set}$? I would have expected $\mathbf{Set} \times \mathbf{Set}$. | |
| Jan 7, 2014 at 16:38 | history | answered | Daniel Schäppi | CC BY-SA 3.0 |