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Objects $X,Y$ with an evaluation morphism $e:X\otimes Y\to I$ and a coevaluation morphism $n:I\to Y\otimes X$ satisfying both snake identities. These are categorical duality data, distinct from the locally convex dual pair.

By the monoidal coherence theorem, calculations may suppress the canonical associators and unitors, restoring them uniquely afterwards. In this notation the two Frobenius monoidal functor identities readWriteWe check both snake identities for this prospective dual pair in a monoidal category. For $FX$, expand and use the first Frobenius identity:The third line uses naturality of $\psi$ and $\phi$. The last line uses $(e\otimes 1)(1\otimes n)=1_X$, followed by the opmonoidal and monoidal unit axioms.

For $FY$, the second Frobenius identity gives the other calculation:Thus both triangular identities hold, andare the evaluation morphism and coevaluation morphism of a dual pair in a monoidal category. Notice that none of the comparison maps was assumed invertible.

Use the specified self-dual-pairing meaning of “coseparable”; categorically this is the Frobenius monoid structure relevant here. The induced monoid object on $FA$ has mapsIts associativity follows from associativity of $m$ and the lax monoidal functor axiom for $\phi$: both iterated products are $F$ of the corresponding triple product, preceded by the same coherent iterated tensor comparison. The unit laws follow from those of $j,m$ and the two monoidal unit axioms. Thus $(FA,J,M)$ is a monoid object.

Its pairing is exactly the evaluation transported in the preceding part:If $n:I\to A\otimes A$ is the original coevaluation morphism, the corresponding new one is ${\psi }_{A,A}F(n){\phi }_0$. The preceding Frobenius monoidal functor calculation proves both snake identities. Hence the pairing of the induced monoid remains self-dual, giving the claimed canonical structure on $FA$. The terminology in the paper imposes this duality condition; it does not add a separability or semisimplicity condition.

The algebra must be finite-dimensional, and there is no universal bound on its dimension. Let $e=\epsilon m$ and write its coevaluation morphism as a finite sumFor any $v\in A$, one snake identity givesThus the finitely many $b_i$ span $A$, making it a finite-dimensional vector space. The other snake identity shows that $e$ is nondegenerate in the other variable as well, so the maps to the dual space determined by this pairing are isomorphisms. In the standard algebraic terminology this is a Frobenius algebra.

Every positive finite dimension of a vector space occurs: take $A=k^d$ with coordinatewise multiplication and $\epsilon (a_1,\dots ,a_d)={\sum }_i{a}_i$. The coordinate idempotents $e_i$ give $e(e_i,e_j)={\delta }_{ij}$ and $n(1)={\sum }_i{e}_i\otimes e_i$. If zero unital algebras are allowed, the zero object gives dimension zero too.

The pairing need not make $A$ a semisimple algebra. For example, the dual numbers $k[t]/(t^2)$ with $\epsilon (a+bt)=b$ have pairing matrixand coevaluation morphism $1\mapsto 1\otimes t+t\otimes 1$. This pairing is nondegenerate over every field, despite the nilpotent ideal $(t)$. Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.