most recent 30 from http://mathoverflow.net 2013-06-20T07:44:41Z http://mathoverflow.net/feeds/question/52989 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctions Garlef Wegart 2011-01-23T20:14:33Z 2011-01-23T23:12:05Z

<p>Given a category $\mathcal C$ together with a monad $T$ on $\mathcal C$, we get an adjunction $$\mathcal C^T(T-,-)\cong \mathcal C(-,\mathrm{For}-).$$</p> <p>Is the same true for 2-monads on a 2-category?</p>

http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctions/53004#53004 Todd Trimble 2011-01-23T22:39:51Z 2011-01-23T22:39:51Z

<p>Well, yes. There are several flavors of 2-monad, as explained in this <a href="http://ncatlab.org/nlab/show/2-monad" rel="nofollow">nLab article</a>, but I'm guessing you mean something like a pseudomonad $T$, and the 2-category $C^T$ of pseudoalgebras over $T$ (with the natural isomorphism replaced by a pseudonatural equivalence). The references in the nLab article may have what you need; I'd look specifically at the article by Blackwell, Kelly, and Power (I don't have the article at hand to confirm this). </p>

http://mathoverflow.net/questions/52989/reference-request-2-monads-and-2-adjunctions/53007#53007 Steve Lack 2011-01-23T23:12:05Z 2011-01-23T23:12:05Z

<p>As Todd says, there are several flavours of 2-monad. </p> <p>If you are interested in strict 2-monads, strict algebras for these, and strict morphisms, then yes you have an adjunction (even an enriched adjunction) as usual. </p> <p>If you mean something weaker, then you will have something weaker than an adjunction. In particular, if you are considering pseudomorphisms of algebras - those which preserve the structure only up to (suitably coherent) isomorphism - then you'll have an <em>equivalence</em> between the category of algebra morphisms from a free algebra $TX$ to an algebra $B$ and the category of morphisms in the base 2-category from $X$ to (the underlying object of) $B$. So rather than an adjunction you'll get some sort of biadjunction. See Corollary 5.6 of the Blackwell-Kelly-Power paper for the case of strict monads, strict algebras, and pseudomorphisms, which is in fact the most important case for many purposes.</p> <p>An important aspect of this is that free algebras $TX$ are <em>flexible</em>, which means among other things that any pseudomorphism from $TX$ to $B$ is isomorphic to a strict one. This is false for a general algebra $A$ in place of $TX$.</p>