A PhD thesis I'm reading contains the following statement (page 21):
Consider an algebra $(A, \nabla, \eta)$, where $\nabla$ is multiplication, $\eta$ is unit, which is also a coalgebra $(A, \Delta, \varepsilon)$, where $\Delta$ is comultiplication, $\varepsilon$ is counit. Then the following two statements are equivalent:
- $\Delta$ and $\varepsilon$ are algebra morphisms.
- $\nabla$ and $\eta$ are coalgebra morphisms.
By definition, an algebra morphism $f:A\to B$ is such a linear map that $$f \circ \nabla_{A} = \nabla_{B} \circ (f \otimes f)$$
where $\nabla_{A}, \nabla_{B}$ are multiplications in $A$ and $B$ respectively, and
$$\eta_{B}=f \circ \eta_{A}$$
where $\eta_{A}, \eta_{B}$ are units in $A$ and $B$ respectively (units are treated as maps $\eta: \mathbb{K}\to A$, where $\mathbb{K}$ is the base field of the algebra).
A coalgebra morphism $f:A \to B$ is such a linear map that $$(f \otimes f)\circ \Delta_{A}=\Delta_{B} \circ f$$ and $$\varepsilon_{B} \circ f=\varepsilon_{A}$$ in similar notation.
I have a problem with the claim: if $\eta: \mathbb{K}\to A$ is a coalgebra morphism, we must define the coalgebra structure on $\mathbb{K}$. It is not obvious to me how to do that. Is comultiplication just doubling $a\to a\otimes a$? Then what is the counit? Is it the map that maps everything to 1? Then this map is not linear. Is it the map that maps everything to zero? Then it does not satisfy the counit diagram. 1 to 1, everything else to 0? Not linear again. Maybe comultiplication is different? What is it, then?
Comultiplication is given by the canonical isomorphism $K\to K\otimes K$ which maps $\lambda $ to $\lambda \otimes 1=1\otimes \lambda$ while counit is given by the identity map. It is straightoforward to verify that these definitions satisfy the axioms.