Excellent question! This gets at the heart of what makes Hopf algebras in vector space-like categories so different from similar structures in the category of sets.

As it turns out, a non-trivial example of the structure you've described cannot exist. The axioms you have laid out are strong enough to force the conclusion that and must both be the trivial monoid .

Let's walk through the proof of this.

Proof of Triviality

We are given the following premises:

1. is a monoid with unit .
2. are submonoids of such that .
3. are monoid homomorphisms for .
4. are retractions, i.e., for all .
5. Coassociativity holds, which implies . (As you noted, the idempotency is a direct consequence of (3) and (4), since for any , , so ).
6. There is a map satisfying the antipode-like axioms for all :
   • (A1)
   • (A2)

The proof proceeds in a few short steps.

Step 1: The maps and are "orthogonal".

Let's see how and interact. For any :

• The element is in the image of , which is .
• The element is in the image of , which is .

By the coassociativity property (5), we have . This means this element lies in . By premise (2), . Therefore, for all , we must have: This means the compositions and are the trivial homomorphisms that map everything to the unit .

A key consequence of this is that maps all of to , and maps all of to . Why? Let . Since is a retraction onto , its image is . So there exists some such that . Then . Symmetrically, for any , .

Step 2: The map is trivial on and .

Let's use the antipode axioms (6) on elements from the submonoids.

• Take any . We can choose in the axiom (A1). . Using the retraction property and the result from Step 1, , this becomes: . So, maps every element of to the unit.

• Take any . We can choose in the axiom (A2). . Using and , this becomes: . So, also maps every element of to the unit.

Step 3: The maps and are trivial.

Now we apply what we've learned back to the antipode axioms for a general element .

• Look at axiom (A1): . The element is in . From Step 2, we know that maps any element of to . So, . The axiom simplifies to: . This must hold for all . Therefore, is the trivial homomorphism.

• Look at axiom (A2): . The element is in . From Step 2, we know . The axiom simplifies to: . This must hold for all . Therefore, is the trivial homomorphism.

Conclusion

Since and are the trivial homomorphisms, their images and must be the trivial monoid . This is the trivial case you wished to avoid.

Where to go from here?

The setup is too restrictive. To find interesting examples, one of the premises must be relaxed.

Possibility 1: Relax the disjointness condition . This is the most common path. If you drop this, then Step 1 of the proof fails.

• Example: Let be any group.
  • Set .
  • Set and for all . (This is the standard diagonal map ).
  • The conditions on being homomorphisms and retractions are met.
  • Coassociativity and holds.
  • The antipode axioms become:
  • These are satisfied by letting , the group inverse. This is the structure of a Hopf algebra on a group. You mentioned this case is "trivial" for your purposes, but it's the canonical example of this kind of structure.

Possibility 2: Consider a different structure for A. The proof relies heavily on the fact that for any , and commute in a very strong way when forming the antipode equations. The structure that feels most natural for your setup is a direct product.

Let and be two non-trivial monoids. Let .

• (isomorphic to )
• (isomorphic to )
• These are submonoids and . This fits!
• These are retractions and homomorphisms.
• Coassociativity holds, and as we showed in Step 1, and are the trivial maps to .

So far, so good. This provides a non-trivial monoid with non-trivial satisfying conditions (1) through (5). The problem is the antipode axiom (6).

Let's check it. Let . and . The axiom (A2) is , which means: This implies that must be the left inverse of . But the expression cannot depend on ! This equation must hold for all . This is only possible if is the trivial monoid. A symmetric argument using (A1) shows must also be trivial.

So even this natural candidate structure is forced to be trivial. This confirms that the axioms as a whole are the problem, not just one specific choice of . Your intuition about looking for Hopf-like structures in has led you to a fascinating "no-go" theorem.