Adjunctions: Algebras of the induced monad VS. Coalgebras of the induced comonad. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:15:16Z http://mathoverflow.net/feeds/question/36766 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36766/adjunctions-algebras-of-the-induced-monad-vs-coalgebras-of-the-induced-comonad Adjunctions: Algebras of the induced monad VS. Coalgebras of the induced comonad. Garlef Wegart 2010-08-26T14:00:39Z 2010-08-26T17:41:01Z <p>Given an adjunction, we get a monad on one side and a comonad on the other side. What is the connection between their algebra and coalgebra categories? Are they allways equivalent?</p> <p>The example i have in mind is the starting point of algebraic geometry (or more general: The fundamental theorem of formal concept analysis) ~ the relation between polynomials and points in affine space given by fRx iff f(x)=0 induces an order preserving isomorphism between radical ideals and "algebraic sets".</p> http://mathoverflow.net/questions/36766/adjunctions-algebras-of-the-induced-monad-vs-coalgebras-of-the-induced-comonad/36775#36775 Answer by Todd Trimble for Adjunctions: Algebras of the induced monad VS. Coalgebras of the induced comonad. Todd Trimble 2010-08-26T15:59:50Z 2010-08-26T17:41:01Z <p>No, they are not generally equivalent. Suppose for example $U: C \to D$ is monadic; this means $U$ has a left adjoint $F: D \to C$ such that the canonical comparison functor $C \to Alg(UF)$ is an equivalence, so that $C$ "is" in effect the category of algebras and $U$ is the forgetful functor. Then you'd be asking whether $Coalg(FU)$ is equivalent to $Alg(UF)$, i.e., whether $C$ is equivalent to the category of $FU$-coalgebras over $C$. </p> <p>For a simple example where this fails, take $C = Set^G$ (category of sets equipped with an action of a group $G$), and $U: Set^G \to Set$ the forgetful functor. The left adjoint is $G \times -: Set \to Set^G$, and one may check that the category of coalgebras is equivalent to $Set$. But $Set$ and $Set^G$ are not equivalent (e.g., $Set^G$ has lots of indecomposable objects given by transitive actions). </p> <p>There is a fairly large class where the canonical comparison functor $Coalg(FU) \to Alg(UF)$ is an equivalence, namely when either $U$ is fully faithful or $F$ is fully faithful. This occurs for example when $C$ and $D$ are posets [edit: this obviously includes the algebraic geometry situation you were considering], and for various "completion processes" (such as sheafification, or taking Cauchy completions on the category of metric spaces and contractive maps, among many others). </p> <p>One might inquire whether or under what circumstances adjunctions $F \dashv U$ lift to adjunctions $Coalg(FU)$ and $Alg(UF)$. This ought to be easy to investigate, but just now my children are rattling around and I'm finding it difficult to concentrate. :-) </p> <p>Edit, now that my children have quieted down: there seems to be no reason for there to be an adjunction between $Coalg(FU)$ and $Alg(UF)$ in general: one can write down various functors in either direction, but I don't see any adjoint pairs cropping up this way. </p> <p>One interesting situation where there is an equivalence between a category of coalgebras and a category of algebras is where a monad has a right adjoint. One may then exhibit a canonical comonad structure on this right adjoint, such that its category of coalgebras is canonically equivalent to the category of algebras of the original monad. This comes up in topos theory, for example, where for a category of presheaves $Set^{C^{op}}$, the evident forgetful functor $Set^{C^{op}} \to Set^{Ob(C)}$ is both monadic and comonadic. </p>