Timeline for Can one compare monads arising from homotopy equivalent adjunctions?
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| Mar 19, 2013 at 0:36 | comment | added | Dylan Wilson | Also- I'm starting to doubt whether $\alpha_R$ comes for free with $\alpha_L$, satisfying the conditions that we want. I think we want what MacLane calls "conjugate natural transformations." See Theorem 2, Section 7, Chapter IV or Categories for the Working Mathematician | |
| Mar 19, 2013 at 0:33 | comment | added | Dylan Wilson | I'll add in details for the map to be multiplicative a little later. For the uniqueness statement I guess I don't know a reference, but the proof is easy: Given $L$ I can recover $R$ by looking at the functor $\text{Hom}(L(-), (-))$ and vice versa. This statement makes sense in quasi-category land but now hom-sets are replaced by hom-spaces and "recover" has the usual "up to a contractible choice" caveat. | |
| Mar 18, 2013 at 19:23 | comment | added | Gregory Arone | Thanks. I would be happy to see a little more details of the argument that the map that you constructed is multiplicative. Do you have a reference to the statement that "adjoint pairs are unique"? | |
| Mar 18, 2013 at 14:30 | comment | added | Dylan Wilson | I should note that proving the result in the infinity-category language is strong enough (in good cases) to give the result about model categories. The above showed that the $\infty$-categories of $T$-algebras and $T'$-algebras are equivalent, and if both of these were, say, combinatorial model categories, then it would follow that the model categories are connected by a zig-zag of Quillen equivalences (I think you need only one zig and one zag, but I haven't convinced myself of this yet.) | |
| Mar 18, 2013 at 14:26 | history | answered | Dylan Wilson | CC BY-SA 3.0 |