Free monad or monad defined from an adjunction. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:16:37Z http://mathoverflow.net/feeds/question/25588 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25588/free-monad-or-monad-defined-from-an-adjunction Free monad or monad defined from an adjunction. unknown (yahoo) 2010-05-22T16:22:19Z 2013-01-13T21:43:31Z <p>My first question here. </p> <p>Accordingly to M. Barr "Coequalizers and free triples" by a free triple (or free monad) generated by an endofunctor $R: X\rightarrow{X}$ we mean a triple $T=(T,\eta,\nu)$ and a natural transformation $p: R\rightarrow{T}$ such that if $T'= (T',\eta',\nu')$ is another triple and $p': R\rightarrow{T'}$ is a natural transformation, then $p'=\tau{p}$, where $\tau:T\rightarrow{T'}$ is a map of triples.</p> <p>On the other hand we have always a monad in the form $(GF,\eta,G\epsilon{F})$ when we have an adjunction $F\dashv{G}$ where $\eta$ is the unit and $\epsilon$ de counit of the adjunction.</p> <p>Can the endofunctor $R$ have the form of an adjunction (for example between a <em>free</em> and a <em>forgetful functor</em>) in the first definition? When does it happen and which is the relationship between the adjunction and the free monad after all? </p> http://mathoverflow.net/questions/25588/free-monad-or-monad-defined-from-an-adjunction/118756#118756 Answer by David White for Free monad or monad defined from an adjunction. David White 2013-01-12T19:46:03Z 2013-01-13T21:43:31Z <p>There seems to be some confusion here regarding language. In your first paragraph, you define the free monad generated by an endofunctor $R$, i.e. one first fixes $R$ then gets the monad. A good toy example for such $R$ would be $U\circ F$ where $F:X\to Y$ is the free functor into some category $Y$ which is well understood and $U:Y\to X$ is the forgetful functor. So the answer to your first question is yes, it can take this form and that's a nice case to play with. But in general $R$ need not take this form. Theorem 5.5 in the paper you cite gives a sufficient condition on an endofunctor $R$ so that free monad on $R$ exists. That condition has nothing to do with $R$ coming from an adjunction.</p> <p>I can interpret your second question in a couple of ways. One way is as "when is $T$ of the form $U\circ F$?'' This is trivial; it's well-known that a monad $T$ is always of the form $U\circ F$ (see <a href="http://en.wikipedia.org/wiki/Monadic_functor#Monads_and_adjunctions" rel="nofollow">wikipedia</a>), often for more than one choice of $U,F$. A better way to interpret your question is "given a monad $T$, how do I determine if it's the free monad generated by some $R$ which takes the form $U\circ F$?" This appears to be a non-trivial problem. Theorem 5.4 in the paper gives one situation where you can determine that your monad is free and recover $R$, but this doesn't classify all such situations. Anyway, even in this nice case where you get $R$ in hand, there are plenty of times such $R$ could fail to be in the form $U\circ F$. For instance, <a href="http://mathoverflow.net/questions/335/is-every-functor-a-composition-of-adjoint-functors" rel="nofollow">this MO answer</a> shows that any such $R$ would need to be a homotopy equivalence on the nerve of $X$. It should not be too hard to construct an endofunctor for which this fails, since there are plenty of self-maps of simplicial complexes which are not homotopy equivalences.</p> http://mathoverflow.net/questions/25588/free-monad-or-monad-defined-from-an-adjunction/118805#118805 Answer by Buschi Sergio for Free monad or monad defined from an adjunction. Buschi Sergio 2013-01-13T12:31:07Z 2013-01-13T12:31:07Z <p>I put this idea:</p> <p>a monad in a (small) category $\mathcal{C}$ is equivalent to a 2-funtor $Mnd\to Cat$ form the free monad $Mnd$ to the 2-category $Cat$. A concrete description of Mnd is: it has one object $\ast$, the hom-category Mnd(0, 0) is the category $\Delta$ of finite ordinals and order-preserving functions, and composition $\Delta \times \Delta\to \Delta$ is ordinal sum. We have the 2-subcategory $L\subset \Delta$ ($L$ for loop) with the unique object $\ast$, full as subcategory, locally discrete.THis inclusion induce the forgetful functor $U: (T, \mu, \eta)\mapsto T$. Now the Left-Kan-extention (as Cat-enriched functor i.e. 2-functor) give the left adjoint of $U$ or in poor words the free-monad of a endofunctor. </p>