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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.

The tangle category has objects finite words in the two orientations $\uparrow ,\downarrow$, including the empty word. A morphism is an oriented compact one-dimensional manifold, embedded in a three-dimensional slab, whose boundary is the prescribed ordered set of top and bottom endpoints. Closed components are allowed. Two such embeddings represent the same morphism when they are ambient-isotopic relative to the boundary.

Read diagrams from top to bottom here, matching the maps drawn in part (d). Composition stacks diagrams with exactly matching intermediate boundary words; horizontal juxtaposition is the tensor product. The empty word is its unit, and straight oriented strands are identities. Crossings, the two oriented evaluations and the two oriented coevaluations generate all morphisms: place an embedding in generic position and slice between its crossings and local extrema. The Reidemeister moves and the cup/cap isotopies describe changes of diagrams without changing the underlying tangle. This is the ordinary unframed version; a framed tangle category instead retains framing data and has a modified first-move rule.

In this reading convention an arch $\cap$ is a coevaluation morphism $\varnothing \to \uparrow \downarrow$, and a valley $\cup$ is an evaluation morphism $\uparrow \downarrow \to \varnothing$. An overleft arrow reverses these strand orientations. We use $X_{+}$ for the crossing whose upper-left-to-lower-right strand passes over the other, and $X_{-}$ for its inverse; this geometric specification avoids ambiguity in sign conventions.

Horizontal juxtaposition is the tensor product in the tangle category. The first panel gives a representative of the first expression: from left to right it consists of the upward identity, the ordinary evaluation morphism, the downward identity, the reversed coevaluation morphism, and the negative crossing. The two arcs are disjoint from the other strands and the crossing is the only over/under intersection in that panel.

The second expression has a boundary-count defect as printed. Under the top-to-bottom convention fixed in part (a), $X_{+}\otimes \uparrow \otimes \uparrow$ has four input and four output endpoints, whereas $\downarrow \otimes \cap \otimes \uparrow$ has only two input and four output endpoints. Those successive boundary words cannot be glued. Reading bottom-to-top does not fix it: the middle factor then has four inputs and two outputs, but the final $\cup$ has no inputs. Thus there is no representative of that literal composite in the tangle category.

An explicit composable correction replaces the middle arch by an evaluation morphism and orients both evaluations to match the strands. Write $e_{\downarrow \uparrow }=\overleftarrow{\cup }$. One such corrected morphism isIts input word is $\downarrow \downarrow \uparrow \uparrow$. The crossing exchanges the first two downward strands, the middle evaluation removes strands two and three, and the outer evaluation removes the remaining downward/upward pair. The second panel is its original drawing. It has four top endpoints and no bottom endpoints. This is a qualified correction, not a claim that the ill-typed printed composite defines a morphism.

Let $n:k\to V^{\ast }\otimes V$ be the supplied coevaluation morphism, $n(1)=x\otimes u+y\otimes v$, and let $e:V^{\ast }\otimes V\to k$ be the supplied evaluation morphism with diagonal coefficients $A,B$. The tangle obtained by inserting the given crossing between the created strand and a through strand, then evaluating, is a removable curl. The first Reidemeister move gives the identityWe calculate this map on both basis vectors. The crossing matrix givesFor input $u$, applying $n$, then $c$, then $e$ givesFor input $v$, the off-diagonal evaluation vanishes and the result is $q^{-1}Bv$. Therefore the removable-curl identity requiresSolve the first equation to obtain $B=q$; substitution in the second gives $A=q^{-1}$. HenceFinally the closed identity strand is represented by $e\circ n$. Its value is $e(x\otimes u+y\otimes v)=A+B$, so the quantum dimension isIt is a weighted categorical contraction, rather than the ordinary vector space dimension $2$. Using a framed category without imposing the removable-curl identity would require additional twist data; the ordinary tangle category and the specified crossing normalization determine these coefficients.

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.

Closing an identity strand using compatible evaluation morphisms and coevaluation morphisms gives the categorical dimension in a pivotal or ribbon setting. For the standard two-dimensional quantum enveloping algebra of sl2 module, the weighted contraction has diagonal coefficients $q^{-1},q$, giving quantum dimension $q+q^{-1}=[2{]}_q$. The chosen duality and framing conventions matter.