A Frobenius functional determines a unique Frobenius coproduct

Choose a basis $(u_1,\dots ,u_n)$ of $A$. The map $\mathrm{\Theta }:A\to A^{\ast }$ given by $\mathrm{\Theta }(v)(a)=ϵ(av)$ is an isomorphism by [F1]. Define $v_i:={\mathrm{\Theta }}^{-1}(u_i^{\ast })$. Then $ϵ(u_j{v}_i)={\delta }_{ji}$, and [F2] gives $a={\sum }_iϵ(a{v}_i)u_i,a={\sum }_i{v}_iϵ(u_{i}a)\mathrm{.}$ For the second identity, apply $z\mapsto ϵ(u_{j}z)$ to its proposed right side: ${\sum }_iϵ(u_j{v}_i)ϵ(u_{i}a)=ϵ(u_{j}a)$. These functionals separate elements by nondegeneracy in the second variable, so the right side equals $a$. Set $e:={\sum }_i{v}_i\otimes u_i$.

Conversely, suppose $(A,\mu ,\eta ,\mathrm{\Delta },ϵ)$ satisfies [F3], and put $d:=\mathrm{\Delta }\circ \eta :k\to A\otimes A$. With the canonical associators and unitors, the two snake composites are $S_L:={\rho }_A\circ (1_A\otimes \beta )\circ {\alpha }_{A,A,A}\circ (d\otimes 1_A)\circ {\lambda }_A^{-1}:A\to A$ and $S_R:={\lambda }_A\circ (\beta \otimes 1_A)\circ {\alpha }_{A,A,A}^{-1}\circ (1_A\otimes d)\circ {\rho }_A^{-1}:A\to A\mathrm{.}$ For $S_L$, replace $\beta$ by $ϵ\mu$ and use the Frobenius identity $(1_A\otimes \mu )\alpha (\mathrm{\Delta }\otimes 1_A)=\mathrm{\Delta }\mu$; naturality of $\alpha$ moves $\eta$ to the first input. The left unit equation reduces the composite to ${\rho }_A(1_A\otimes ϵ)\mathrm{\Delta }$, which is $1_A$ by the right counit. For $S_R$, use $(\mu \otimes 1_A){\alpha }^{-1}(1_A\otimes \mathrm{\Delta })=\mathrm{\Delta }\mu$; the right unit equation reduces it to ${\lambda }_A(ϵ\otimes 1_A)\mathrm{\Delta }$, which is $1_A$ by the left counit. Thus both snake composites are identities.

The first displayed dual-basis identity gives ${\sum }_iϵ(a{v}_i)u_i=a$, and the second gives ${\sum }_i{v}_iϵ(u_{i}a)=a$, so this $e$ has the required two contraction properties. If another tensor has the first property, contracting its first factor against every functional $x\mapsto ϵ(ax)$ gives the same vector for every fixed $a$; these functionals range over all of $A^{\ast }$ by nondegeneracy. Hence the tensors are equal. The contraction map $A\otimes A\to {\mathrm{Hom}}_k(A^{\ast },A)$ is an isomorphism in finite dimension, so the tensor is intrinsically defined and independent of the basis.

To prove $(a\otimes 1)e=e(1\otimes a)$, test the two tensors against arbitrary $\varphi \otimes \psi \in A^{\ast }\otimes A^{\ast }$. Nondegeneracy supplies $c,d\in A$ with $\varphi (z)=ϵ(cz)$ and $\psi (z)=ϵ(zd)$. The first contraction is ${\sum }_iϵ(ca{v}_i)ϵ(u_{i}d)=ϵ(cad)$ by the second dual-basis identity applied to $d$. The second is ${\sum }_iϵ(c{v}_i)ϵ(u_{i}ad)=ϵ(cad)$ by the same identity applied to $ad$. Since such product functionals separate tensors, the two tensors agree.

Define $\mathrm{\Delta }(a)=(a\otimes 1)e$. Then $(ϵ\otimes 1)\mathrm{\Delta }(a)={\sum }_iϵ(a{v}_i)u_i=a$. Also $(1\otimes ϵ)\mathrm{\Delta }(a)=a{\sum }_i{v}_iϵ(u_i)=a$ by the second identity in step 1.1 at $a=1$. The central relation of step 2.2 gives $\mathrm{\Delta }(ab)=(a\otimes 1)\mathrm{\Delta }(b)=\mathrm{\Delta }(a)(1\otimes b)\mathrm{.}$ Expanding these two equalities with the associator is exactly the pair of Frobenius identities in [F3].

The defining contraction also yields, for all $x,y,a\in A$, $(\beta \otimes \beta )(x\otimes y,\mathrm{\Delta }(a))={\sum }_iϵ(xa{v}_i)ϵ(y{u}_i)=ϵ(yxa)=\beta (yx,a)\mathrm{.}$ This establishes the contraction identity.

If ${\mathrm{\Delta }}^{\mathrm{\prime }}$ is another coproduct with counit $ϵ$ and the two Frobenius identities, write $e^{\mathrm{\prime }}={\mathrm{\Delta }}^{\mathrm{\prime }}(1)={\sum }_j{x}_j^{\mathrm{\prime }}\otimes y_j^{\mathrm{\prime }}$. Applying the counit identities to ${\mathrm{\Delta }}^{\mathrm{\prime }}(a)$ and using each Frobenius identity gives ${\sum }_jϵ(a{x}_j^{\mathrm{\prime }})y_j^{\mathrm{\prime }}=a,{\sum }_j{x}_j^{\mathrm{\prime }}ϵ(y_j^{\mathrm{\prime }}a)=a\mathrm{.}$ Step 2.1 therefore gives $e^{\mathrm{\prime }}=e$. The Frobenius identity then gives ${\mathrm{\Delta }}^{\mathrm{\prime }}(a)=(a\otimes 1)e^{\mathrm{\prime }}=(a\otimes 1)e=\mathrm{\Delta }(a)$, so the coproduct is unique.

Use the tensor-product pairing ${\beta }^{\otimes 3}$ to test coassociativity. For any $x,y,z,a\in A$, step 4.1 gives $⟨x\otimes y\otimes z,(\mathrm{\Delta }\otimes 1)\mathrm{\Delta }(a)⟩=\beta (zyx,a)=⟨x\otimes y\otimes z,(1\otimes \mathrm{\Delta })\mathrm{\Delta }(a)⟩\mathrm{.}$ The pairing ${\beta }^{\otimes 3}$ is nondegenerate: in any basis its matrix is the Kronecker product of three invertible matrices for $\beta$. Thus the two tensors are equal, proving coassociativity.

The two snake identities exhibit inverse maps between $A$ and $A^{\ast }$ for the left and right pairings, respectively; hence $\beta$ is nondegenerate in both variables. It follows that $ϵ$ is a Frobenius functional by [F1], and step 4.2 gives uniqueness of $\mathrm{\Delta }$. Symmetry of $\beta$ and commutativity of $\mu$ remain the separate properties defined in [F3].