# All about (co)algebras for the identity functor

Any fact about these would be of interest. Has anyone seen an interesting structure constructed from a monad and comonad based on a single identity functor?

I am working on a very small example just so that I can do all the calculations. In particular, suppose I have a category with one object that's a set with two elements. The morphisms are all endo-functions of this set. Now take the identity endo-functor and use it to generate a monad and comonad. Next, find the (co)algebra for these. Are these completely trivial?

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Does anyone want to say definitively that there are no interesting (co)monads based on identity functor? –  Ben Sprott Jun 4 '12 at 18:47

Let's unwind the definition. A monad on the identity functor is just a pair of endomorphisms $\eta$ and $\mu$; the defining identities for a monad are automatically defined.

An algebra for this monad is an endomorphism $h$ such that:

• $h \circ h = \mu \circ h$
• $\eta \circ h = 1$

In your problem, the endomorphisms form a group, and so there is a unique algebra $h = \mu$ if $\eta \circ \mu = 1$, and no algebras otherwise.

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Hey, I can't vote your answer up yet as i have no reputation points. In lieu of that: Thanks! –  Ben Sprott Jun 3 '12 at 19:55
@Ben: I think you should still be able to click the check mark, if you want to accept this as the answer. –  Hurkyl Jun 5 '12 at 0:01
I gave it the check. Thanks again. –  Ben Sprott Jun 5 '12 at 0:36