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So many times you can write out the axioms for an algebraic structure (say an algebra over a ring) as commutative diagrams and then reverse all the arrows and get a new structure: say a coalgebra. Can we do this with properties of individual elements? Specifically - is there a notion of a co-idempotent element of a coalgebra?

I tried defined the set of idempotents of an algebra $A$ as the set $I \subset A$ for which $d \circ \mu$ is the identity, where $d(a) = a\otimes a$ and $\mu$ is the multiplication map. To dualize this, said the set of coidempotents is the one such that the map $\Delta \circ b$ is the identity, but this just begged me to find a co-version $b$ of the map $d(a) = a\otimes a$ - which was a bit silly.

I know there's a notion of idempotent and coidempotent modules of a ring - though I don't know how this relates to individual elements.

Is there a notion of a co-idempotent? How can I reduce the idempotent element property to a commutative diagram and reverse the arrows to come up with it? Is this even possible?

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  • $\begingroup$ The problem is that the notion of an idempotent is a quadratic property, hence not straightforward to dualize. $\endgroup$ – darij grinberg Feb 25 '14 at 2:38
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Idempotents make sense in a monoid, not just in an algebra; their definition doesn't refer to the addition at all. For a monoid $M$ with multiplication $\mu : M \times M \to M$, the set of idempotents is the equalizer of the identity $\text{id}_M : M \to M$ and the map $\mu \circ \Delta : M \to M$, where $\Delta$ is the diagonal $\Delta(m) = (m, m)$.

So what's the deal with $\Delta$? The existence of $\Delta$ reflects the fact that any set is canonically a comonoid with respect to $\times$, with comultiplication given by $\Delta$. (This is true in any category with finite products; see this blog post for details.) So monoids in $\text{Set}$ are canonically always bimonoids (and this is why taking the free vector space on a monoid in $\text{Set}$ gives you a bialgebra, not just an algebra), and for any bimonoid $M$ it makes sense to consider the composition $\mu \circ \Delta$ where $\Delta$ is the comultiplication and $\mu$ the multiplication, so you can take its equalizer with the identity. The dual notion, I guess, is the coequalizer (note that the dual of a subobject is a quotient object and that the dual of a bimonoid is the same bimonoid, but regarded as an object in the opposite category and with the multiplication and comultiplication trading roles.)

When working with algebras one would like to think of them as monoids in $\text{Vect}$ with respect to $\otimes$, but since $\otimes$ is not the cartesian product in $\text{Vect}$ the above argument doesn't apply and it's no longer natural to write down the map $\Delta$; indeed, note that $\Delta$ is not linear. So in this context it's not clear to me in what sense we can dualize the idea of idempotents. (The natural home for the notion of idempotents is really in a category, and I don't know anything about cocategories.)


When we try to dualize the story for just monoids, we get the following. The notion dual to a diagonal map $\Delta : M \to M \times M$ in a category with finite products is a codiagonal map $\nabla : M \sqcup M \to M$ in a category with finite coproducts. If $C$ is such a category, then codiagonal makes every object in $C$ canonically a monoid with respect to coproduct, so any comonoid in $C$ (with respect to the coproduct) is canonically a bimonoid and in particular one can write down the composition $\nabla \circ \Delta$ where $\Delta$ is the comultiplication, then take its coequalizer.

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An idempotent in a ring is the same thing as an idempotent in the endomorphism ring of its regular left module. You can define, if you want, an idempotent of a coalgebra to be an idempotent endomorphism of its regular left comodule.

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