Monoidal fibres, bialgebras and duality reconstruction

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

An exact faithful finite-dimensional observation functor recovers a coalgebra. Its tensor comparisons recover multiplication. Evaluation and coevaluation then determine an antipode, and the two directions of duality determine whether that antipode is invertible. We construct each structure and prove the converse, keeping the order of right-comodule coefficients explicit.

Fix a field \(k\), an essentially small \(k\)-linear abelian monoidal category \(A\), and an exact faithful \(k\)-linear strong monoidal functor \(F:A\to\operatorname{Vect}^{\rm fd}_k\). Its comparisons have direction

All sums, diagrams and categories of representatives are small in fixed universes. Parentheses in vector tensors use their usual associativity maps. No permutation of factors in \(A\) is implicit.

Coend coalgebras and reconstruction from finite fibres proves the full underlying equivalence with finite right comodules, including its actual Ind inverse and finite-part argument. Write \(H=C_F\) for that coefficient coalgebra. For \(v\in F(a)\), \(\phi\in F(a)^*\), and \(f:a\to b\), its defining relations and coproduct are

Here \(e_i,e_i^*\) are any dual bases of \(F(a)\). The canonical coaction is \(\rho_a(v)=\sum_i e_i\otimes[e_i^*\otimes v]_a\).

Tensor duality, Sections 2–3 proves both triangle identities and the uniqueness of chosen duals. Strict monoidal models, Sections 1–3 supplies the ordered coherence convention; the structural maps below retain their indicated domains. Completed tensors, Section 5 treats duality of already specified bialgebra structures. The construction from the fibre functor is proved here. Classical reconstruction context is [EGNO], Sections 5.3–5.4.

We use Hopf algebra to mean a bialgebra with an antipode satisfying both convolution equations. When its antipode is required to be bijective, we say so explicitly.

1. Multiplication and unit from the strong comparisons

For two coefficients define

The inverse tensor comparison in the functional and the forward comparison in the vector are both part of this formula. Let \(u:k\to H,\ \lambda\mapsto\lambda1_H\), be the unit map.

Lemma 1.1. This descends to a unital associative multiplication on \(H\).

Proof: relations. For \(f:a\to a'\), naturality gives \(F(f\otimes1)J_{a,b}=J_{a',b}(F(f)\otimes1)\). Apply relation (0.2) to \(f\otimes1_b\), using that equality for its vector and inverse equality for its functional. It identifies the two products obtained from \([\phi'\otimes F(f)v]_{a'}=[\phi'F(f)\otimes v]_a\). The same argument with \(1_a\otimes g\) handles every relation in the second input. Bilinearity therefore gives the product on the quotient.

Proof: associativity. For three inputs, let \(J_L\) combine them first in \(a,b\), then with \(c\); let \(J_R\) combine \(b,c\) first. The strong monoidal coherence equation is

with the ordinary vector associator understood. The two iterated products have vectors \(J_L(v\otimes w\otimes z)\), \(J_R(v\otimes w\otimes z)\), and functionals \((\phi\otimes\psi\otimes\chi)J_L^{-1}\), \((\phi\otimes\psi\otimes\chi)J_R^{-1}\). Relation (0.2) for the actual associator identifies them. Thus the products agree on all spanning coefficients.

The strong unit equations say that \(F(l_a)J_{\mathbf1,a}(j_0\otimes1)\) and \(F(r_a)J_{a,\mathbf1}(1\otimes j_0)\) are the standard vector unit maps. Apply relation (0.2) to those unitors in (1.1). Both products with the specified \(1_H\) are the original coefficient. This proves both unit laws. \(\square\)

Proposition 1.2. With (0.2), this is a bialgebra.

Proof. For coefficients \(x=[\phi\otimes v]_a\), \(y=[\psi\otimes w]_b\), choose bases \(e_i,f_j\). A basis of \(F(a\otimes b)\) is \(J_{a,b}(e_i\otimes f_j)\), with the corresponding inverse-comparison dual functionals. Substitution in (0.2) gives

This is exactly \(\Delta(x)\Delta(y)\) in the tensor-product algebra, with the middle vector-space interchange included. The counit of (1.1) is \(\phi(v)\psi(w)=\epsilon(x)\epsilon(y)\). Because \(F(\mathbf1)\) is one-dimensional with basis \(j_0(1)\), its unit coefficient satisfies \(\Delta(1_H)=1_H\otimes1_H\) and \(\epsilon(1_H)=1\). These are all compatibility equations. \(\square\)

2. The monoidal equivalence and its morphisms

For any bialgebra \(H\), right comodules have the tensor coaction

Every displayed Sweedler sum is a finite tensor expression on each vector. Coassociativity follows by applying the two original coaction equations and \(\Delta(xy)=\Delta(x)\Delta(y)\). The counit follows from its multiplicativity. The unit comodule is \(k\) with \(1\mapsto1\otimes1_H\). Associativity and both unit identifications are comodule maps by associativity and the unit laws in \(H\); their coherence is that of the underlying vector maps. Tensoring two comodule maps respects (2.1) by their individual intertwining equations.

For the reconstructed \(H\), formula (1.1) makes \(J_{a,b}\) a comodule map: use the tensor basis of Proposition 1.2 in the coaction of \(F(a\otimes b)\), and obtain (2.1) transported through \(J_{a,b}\). The unit coefficient makes \(j_0\) a comodule map. Hence the underlying equivalence from the coend lesson is a strong monoidal equivalence, using these very comparisons and all its previously proved maps. Tensor is exact in each variable: the vector tensor is exact over a field, and the exact faithful underlying functor on comodules reflects exactness.

These structures are forced by the requirement. Equating the two coactions for a proposed \(J_{a,b}\) and applying the functional \(\phi\otimes\psi\) to the first tensor factor recovers the coefficient product (1.1). The unit comparison similarly forces \(1_H\). Since coefficients span, no other multiplication or unit can satisfy the same requirements.

Conversely begin with a bialgebra \(H\) and its finite comodule category with its usual forgetful functor and (2.1). The coefficient isomorphism of the coend lesson sends \([\phi\otimes m]_M\) to \((\phi\otimes1)\rho_M(m)\). Formula (2.1) makes it respect multiplication and unit, so it recovers the original bialgebra, not just its coalgebra. This proves reconstruction up to equivalence preserving the strong fibre functor and up to bialgebra isomorphism.

For explicit functoriality, let \(G:A\to A'\) be a strong monoidal \(k\)-linear functor and let \(\theta:F'G\xrightarrow{\sim}F\) be a monoidal natural isomorphism. It induces

Naturality of \(\theta\) respects every coefficient relation, so this is a coalgebra map. The commuting tensor-comparison square for a monoidal \(\theta\) and the unit square make (2.2) respect (1.1). Thus it is a bialgebra map. Identities and composites follow on these coefficients. No claim that an arbitrary underlying natural isomorphism respects multiplication is being made.

3. Left duals give the antipode coefficient map

Our left dual \(a^\vee\) has evaluation \(\operatorname{ev}_a:a^\vee\otimes a\to\mathbf1\) and coevaluation \(\operatorname{coev}_a:\mathbf1\to a\otimes a^\vee\), with both triangle identities. The actual transported maps are

They are a pairing \(F(a^\vee)\otimes F(a)\to k\) and a coevaluation tensor in \(F(a)\otimes F(a^\vee)\). Strong monoidal coherence transports the two specified triangles to the two vector-space triangles.

They identify \(F(a^\vee)\) with \(F(a)^*\). Explicitly put

For completeness, if the transported coevaluation is \(\sum_s v_s\otimes u_s\), its inverse sends \(\phi\) to \(\sum_s\phi(v_s)u_s\). The first triangle makes \(d_ad_a^{-1}\phi=\phi\), and the second makes \(d_a^{-1}d_a(u)=u\). This proves nondegeneracy without assuming a basis pairing beforehand.

For \(f:a\to b\), its dual \(f^\vee:b^\vee\to a^\vee\) is obtained by inserting the coevaluation of \(a\), applying \(1\otimes f\otimes1\), then evaluating \(b^\vee\otimes b\). Both triangles show that its evaluation equation is

That equation characterizes its fibre map; the ordinary dual-map uniqueness in \(A\) follows from the two triangle factorizations. In particular identities and reversed composites give the actual contravariant dual functor.

Suppose every object has a left dual. For \(v\in F(a)\), let \(\widehat v:F(a)^*\to k\) be evaluation at \(v\). Define

This respects the relation for every \(f:a\to b\). Indeed (3.2) identifies \(\widehat{F(f)v}d_b\) with \(\widehat v\,d_aF(f^\vee)\); relation (0.2) for \(f^\vee\) moves that map into the vector factor, giving \(d_a^{-1}(\phi F(f))\) there. These are exactly the two images of the original relation. Thus (3.3) gives a well-defined linear map \(S:H\to H\).

4. Both antipode equations and the opposite laws

In a basis of \(F(a)\), put \(h_{ij}=[e_i^*\otimes e_j]_a\). Identify the dual fibre by \(d_a\), and let \(k_{ij}\) be its coefficient matrix in the dual basis. Formula (3.3) says \(S(h_{ij})=k_{ji}\). Since evaluation and coevaluation are comodule maps, their coefficients give respectively

The first follows by evaluating the coaction of \(e_i^*\otimes e_j\); the second by comparing the coefficients of \(\sum_r e_r\otimes e_r^*\). With the stated transpose these are

Since all these coefficients span \(H\), (4.2) proves both antipode equations \(S*\operatorname{id}_H=\operatorname{id}_H*S=u\epsilon\).

Here convolution of maps from a coalgebra to an algebra means \(p*q=m(p\otimes q)\Delta\); its associativity and unit follow by expanding the two triple sums and using the coassociative coproduct and associative product. Two two-sided inverses in any such algebra are equal: \(p=p*(f*q)=(p*f)*q=q\). Thus the antipode is unique. Changing the choices of left duals cannot change it.

Lemma 4.1. The antipode satisfies

Proof: multiplication. In the convolution algebra of maps \(H\otimes H\to H\), the map \(S\circ m\) is inverse to \(m\), by multiplicativity of the coproduct and both antipode equations. The map \((x,y)\mapsto S(y)S(x)\) is another inverse. For one composite its sum is \(\sum S(y_1)S(x_1)x_2y_2=\epsilon(x)\epsilon(y)1_H\); for the other use \(\sum x_1y_1S(y_2)S(x_2)\) and cancel the middle \(y\)-pair first. Uniqueness proves the product identity. Evaluating the antipode equation at the grouplike \(1_H\) gives \(S(1_H)=1_H\).

Proof: coproduct. In convolution of maps \(H\to H\otimes H\), the map \(\Delta S\) is inverse to \(\Delta\), because \(\Delta\) is an algebra map. A second inverse is \(T_0=(S\otimes S)\tau\Delta\). In a fourfold coassociative expansion, its two composites with \(\Delta\) are

In the first, cancel \(S(h_2)h_3\), then the remaining \(S(h_1)h_2\); in the second cancel \(h_2S(h_3)\), then \(h_1S(h_2)\). Both become \(\epsilon(h)(1_H\otimes1_H)\). Uniqueness gives the coproduct identity. Applying \(\epsilon\) to the antipode equation gives \((\epsilon S)*\epsilon=\epsilon\) in the scalar convolution algebra, whose unit is \(\epsilon\). Hence \(\epsilon S=\epsilon\). \(\square\)

5. Right duals give an actual inverse antipode

A right dual \(a^r\) has evaluation \(a\otimes a^r\to\mathbf1\) and coevaluation \(\mathbf1\to a^r\otimes a\). Its transported pairing identifies \(F(a^r)\) with \(F(a)^*\), just as in Section 3 but with the factors in this specified order.

Write \(d_a^r:F(a^r)\xrightarrow{\sim}F(a)^*\) for the right-pairing identification. Its inverse is obtained by contracting the right coevaluation, with the two inverse identities proved by the two right triangles. For \(f:a\to b\), insert the coevaluation of \(a\) on the left of \(b^r\), apply \(1\otimes f\otimes1\), and evaluate \(b\otimes b^r\). This gives \(f^r:b^r\to a^r\) and the identity \(d_a^rF(f^r)=F(f)^*d_b^r\).

If every object has such a right dual, define

For a defining coefficient relation, the preceding dual-map identity moves \(F(f^r)\) from the functional to the vector by (0.2), giving exactly the two images under this formula. Thus it descends, with every variance specified. Let the right-dual coefficient matrix be \(r_{ij}\), so \(T(h_{ij})=r_{ji}\). Evaluation and coevaluation in their right-dual order now give

For example the first evaluates \(e_j\otimes e_i^*\); the second compares the coefficients of \(\sum_p e_p^*\otimes e_p\). Thus

These are the two antipode equations for the co-opposite bialgebra \(H^{\rm cop}\), with coproduct \(\tau\Delta\). That is a bialgebra: the flip is an algebra map from the tensor-product algebra to itself, and the two bialgebra compatibility equations are preserved. Hence Lemma 4.1 also makes \(T\) anti-multiplicative.

Suppose both left and right duals exist. Apply the anti-multiplicative \(S\) to the two opposite equations for \(T\). In ordinary convolution the resulting equations say that \(S\circ T\) is a two-sided inverse to \(S\). The identity map is already such an inverse by (4.2). Uniqueness therefore gives \(S\circ T=\operatorname{id}_H\).

Apply the anti-multiplicative \(T\) to the ordinary equations for \(S\). In opposite convolution \(p*_{\rm cop}q=m(p\otimes q)\tau\Delta\), the resulting equations make \(T\circ S\) a two-sided inverse to \(T\). The identity is already that inverse by (5.2). Hence \(T\circ S=\operatorname{id}_H\). This proves actual bijectivity, with

No finite-dimensionality of the entire coalgebra was used. One direction of duality supplies its specified antipode equations; the second direction is what supplied this inverse.

6. Converse duals and ordinary symmetry

Conversely, let \(H\) be a bialgebra with an antipode \(S\), and let \(V\) be a finite right comodule with coefficient matrix \(h_{ij}\). Define a coaction on the ordinary dual basis by the matrix

The anti-coalgebra identity in (4.3) gives \(\Delta k_{ij}=\sum_r k_{ir}\otimes k_{rj}\), and the counit gives \(\epsilon k_{ij}=\delta_{ij}\). Thus this really is a coaction. Equations (4.2) say exactly that the ordinary evaluation \(V^*\otimes V\to k\) and coevaluation \(k\to V\otimes V^*\) are comodule maps. Their ordinary vector triangles are identities, so they give a left dual. Transposing a comodule map respects these coactions by its coefficient equation and linearity of \(S\), hence supplies all dual maps.

If \(S\) is bijective, (4.3) makes \(S^{-1}\) anti-multiplicative and anti-coalgebraic as well. Applying \(S^{-1}\) to the antipode equations in reverse product order gives its two equations for \(H^{\rm cop}\). The matrix \(r_{ij}=S^{-1}(h_{ji})\) therefore gives the right dual, with evaluation \(V\otimes V^*\to k\) and coevaluation \(k\to V^*\otimes V\), by (5.1). Consequently finite comodules have both duals exactly when the reconstructed antipode is bijective; the reverse implication was Section 5 applied to all their fibres. Their coefficients span \(H\), as proved in the coend reconstruction lesson.

For a bialgebra map \(q:H\to H'\) between bialgebras with antipodes, \(qS\) and \(S'q\) are both convolution inverses to \(q\) in \(\operatorname{Hom}(H,H')\), by its coalgebra, algebra and unit equations. Therefore \(qS=S'q\). In particular the map (2.2) preserves antipodes whenever the dual hypotheses supply them.

Finally the ordinary vector flip \(V\otimes W\to W\otimes V\) is a comodule map for all finite comodules if and only if \(H\) is commutative. In the forward direction compare the two tensor coactions and apply a pair of fibre functionals: every coefficient of \(V\) commutes with every coefficient of \(W\). Such coefficients span \(H\), so all products commute. In the reverse direction the two coactions (2.1) agree after flipping because their last coefficients commute.

Thus a symmetry on \(A\) preserved by \(F\) forces commutativity. Conversely commutativity gives the usual flip symmetry on finite comodules, which the strong monoidal equivalence transports to \(A\). Full faithfulness gives its actual symmetry maps; the symmetry and hexagon laws follow because their vector maps satisfy them and \(F\) is faithful. This criterion concerns the ordinary flip; no claim that an arbitrary abstract braiding forces commutativity is involved.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: a monoid grading need not have duals)

Let \(M\) be a small monoid. Give finite-dimensional \(M\)-graded vector spaces the ordered tensor grading \(g\cdot h\). Reconstruct its bialgebra and determine exactly when it has an antipode. Explain the obstruction for the nonnegative-integer grading.

Solution. The underlying graded category and forgetful functor satisfy the hypotheses: kernels and cokernels are grade by grade, and forgetting is exact and faithful. The tensor is the direct sum of \(V_g\otimes W_h\) in grade \(gh\); associativity and the identity grade give its strong ordinary forgetful comparisons.

The graded coefficient calculation in the coend lesson gives the basis \(g_m\), \(m\in M\), with \(\Delta g_m=g_m\otimes g_m\) and \(\epsilon g_m=1\). Formula (1.1) gives \(g_mg_n=g_{mn}\) and \(1_H=g_e\). This is the monoid bialgebra \(k[M]\).

Its grouplike elements are precisely the \(g_m\). Indeed write a grouplike as a finite sum \(\sum a_mg_m\). Comparing its coproduct with its tensor square makes \(a_ma_n=0\) for \(m\ne n\), and \(a_m=a_m^2\). The counit is \(1\), so exactly one coefficient is \(1\). If an antipode exists, (4.3) makes \(S(g_m)\) grouplike, and its two equations make the corresponding monoid element a two-sided inverse to \(m\). Hence every \(m\) is invertible and \(M\) is a group. Conversely for a group, \(S(g_m)=g_{m^{-1}}\) satisfies both equations on each basis vector and is bijective.

For \(M=\mathbb N\) under addition the bialgebra is \(k[t]\) with \(t\) grouplike, rather than primitive. There is no inverse to the degree-one monoid element, so no antipode. A degree-one line cannot have a left or right dual within nonnegative gradings: every tensor with it has grades at least one, so an evaluation to the degree-zero unit is zero and cannot satisfy its triangle. This gives the same obstruction directly in the category.

Exercise 2 (standard: finite groups in every characteristic)

For a finite group \(G\), reconstruct the bialgebra of its finite-dimensional representations with their ordinary tensor and forgetful functor. Compute its antipode, and separate commutativity from cocommutativity without assuming semisimplicity.

Solution. Send a coefficient to the function \(g\mapsto\phi(gv)\). It respects every representation-map relation. Every function occurs from a coefficient of the regular module \(k[G]\), using the vector \(1\) and an arbitrary functional on \(k[G]\). Conversely the module map \(k[G]\to V\), \(r\mapsto rv\), moves any coefficient to the regular module with vector \(1\), just as in the regular-module calculation of the coend lesson. A zero coefficient function then gives the zero functional there; finite sums combine through direct sums. This proves the coalgebra isomorphism \(H=k^G\).

The tensor action gives pointwise multiplication of coefficient functions, and the unit is the constant function \(1\). The coproduct, counit and antipode are

Evaluating the two antipode composites on \(g\) gives \(f(g^{-1}g)=f(e)\) and \(f(gg^{-1})=f(e)\), so both equations hold and \(S^2=1\). Pointwise multiplication is commutative, in accordance with the usual flip symmetry on representations. The coproduct is cocommutative exactly when \(G\) is abelian: if \(gh\ne hg\), the indicator function of \(gh\) distinguishes the two values of the coproduct. These calculations use neither averaging nor a decomposition into irreducibles, and hold when the characteristic divides \(|G|\).

Exercise 3 (advanced: the characteristic-two obstruction)

Let the coalgebra \(C\) have basis \(1,x\), with \(\Delta1=1\otimes1\), \(\Delta x=1\otimes x+x\otimes1\), and counit \(\epsilon1=1,\epsilon x=0\). Determine when the proposed multiplication \(x^2=0\) makes it a bialgebra. Relate this to tensoring two square-zero operators, and determine whether a different unital multiplication could work in characteristic other than two.

Solution. Multiplicativity of the coproduct requires

Since \(x\otimes x\ne0\), this forces characteristic two. In characteristic two the relation is preserved, so the product defines a bialgebra: multiplication on the unit and counit is compatible, and the displayed calculation checks the only defining relation. The map \(S(1)=1,S(x)=-x=x\) satisfies both antipode equations on the basis, and is its own inverse.

For square-zero operators \(N,M\), the primitive tensor prescription is \(N\otimes1+1\otimes M\). Its square is \(2N\otimes M\). Taking both to be nonzero Jordan blocks of size two makes \(N\otimes M\ne0\), so the tensor prescription fails outside characteristic two. In characteristic two it is again square-zero; ordinary tensor associativity and the zero operator on the unit preserve these sums. Tensoring intertwining maps respects them, giving the strong forgetful tensor structure. The ordinary dual has operator \(-N^*\); evaluation and coevaluation commute with these sums, and the vector triangles give both duals.

There is no alternative bialgebra multiplication on this coalgebra outside characteristic two. First its only grouplike element is \(1\): a grouplike with counit one has form \(1+ax\), and comparison of coproduct with tensor square gives \(a^2=0\), hence \(a=0\). Thus a bialgebra unit would have to be this \(1\). Write \(x^2=b1+cx\); multiplicativity of the counit gives \(b=0\). Comparing \(\Delta(x^2)\) with \((\Delta x)^2\) still leaves the term \(2x\otimes x\), so it forces characteristic two regardless of \(c\). This is an obstruction to compatible multiplication, even though the coalgebra and its finite-comodule category exist over every field.

Exercise 4 (synthesis: a valid lax functor is not sufficient)

Let \(B=k\times k\), and set \(F_B(V)=V\otimes B\) on finite-dimensional vector spaces. Give it its lax tensor comparison using multiplication in \(B\) and its lax unit using \(1_B\). Verify that it is exact and faithful but not strong. Compute its coefficient coalgebra and show that this coalgebra admits no bialgebra structure.

Solution. The lax comparison first interchanges the middle \(B\) and vector factors, then multiplies the two \(B\)'s; its unit sends \(1\) to \(1_B\). The associativity and both unit equations are exactly those of the unital algebra \(B\), with ordinary vector reassociation. Thus this is a valid lax monoidal functor. Tensor with the nonzero vector space \(B\) is exact over the field and detects every nonzero map, so it is exact and faithful. Its unit has dimensions \(1\to2\), and its tensor comparison at \(k,k\) has dimensions \(4\to2\); neither is invertible.

Its coefficient coalgebra is the matrix coalgebra \(B^*\otimes B\). To prove the quotient identification, write a vector in \(V\otimes B\) as \(\sum_i v_i\otimes b_i\), and send its coefficient with functional \(\phi\) to

This is independent of the tensor expression by its bilinear universal property, and respects all linear maps on \(V\). Conversely the coefficients at the object \(k\), whose fibre is \(B\), give an inverse: the maps \(k\to V\), \(1\mapsto v_i\), turn the defining coefficient relations into exactly (7.3). Formula (0.2) gives its usual matrix coproduct, so its dual algebra is \(\operatorname{End}_k(B)\simeq M_2(k)\).

A grouplike element of this coalgebra would give, by evaluation, a unital algebra map \(M_2(k)\to k\). Such a map cannot exist. Its values on \(E_{11},E_{22}\) are orthogonal idempotents adding to \(1\), so one is \(1\) and the other \(0\). But the scalar products of the images of \(E_{12},E_{21}\) commute, whereas their products in the two orders are \(E_{11}\) and \(E_{22}\), forcing their images equal. This is a contradiction over every field.

There is therefore no grouplike element, and a bialgebra unit would have to be grouplike with counit one. No bialgebra structure exists on this coalgebra. Exactness, faithfulness and valid lax coherence do not supply the invertible comparisons needed in (1.1).

References

[EGNO] Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych and Victor Ostrik, Tensor Categories, Sections 5.3–5.4, especially Theorem 5.4.1. Freely available author-hosted final version. These sections give classical bialgebra and duality reconstruction context. The coefficient operations, their compatibility proofs, the explicit left/right antipodes and inverse, and the counterexamples used here are given above, with the linked underlying coalgebra reconstruction.