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An algebra over a commutative ring that is also a coalgebra, whose comultiplication and counit preserve the multiplication and unit. Equivalently it is a bimonoid in $k$-modules.

A linear subspace $I$ of a coalgebra satisfying the displayed conditions. These are exactly the conditions for the comultiplication and counit to descend to the quotient vector space $C/I$. An ideal that is also a coideal in a bialgebra gives a quotient bialgebra.

A right comodule is an object $M$ with a coaction $\rho :M\to M\otimes C$ satisfying the coassociativity and counit laws. This definition extends from coalgebras to comonoids in a monoidal category.

The categorical dual of a monoid object: an object with coassociative comultiplication and a counit, with the ambient constraints included.

The tensor of left modules over a bialgebra has action $h(u\otimes v)=\sum (h_{(1)}u)\otimes (h_{(2)}v)$. The unit module $k$ has action through the counit.

Use $q\in k^{\times }$ and the quantum plane convention $yx=qxy$. In this convention $S{L}_q(2)$ is the coordinate Hopf algebra of quantum SL2 with the catalog parameter replaced by $q^{-1}$. It is generated by $a,b,c,d$, subject toEquivalently the last determinant relation is $da-qbc=1$. The quantum determinant convention is essential: these relations use the inverse of the parameter in the catalog's $ab=qba$ convention.

Writing $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the comultiplication is matrix multiplication, $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$. Explicitly,The counit is $\epsilon (a)=\epsilon (d)=1$, $\epsilon (b)=\epsilon (c)=0$. Extend $\Delta$ and $\epsilon$ as unital algebra homomorphisms over a field. Matrix multiplication is associative, giving coassociativity; multiplication by the identity matrix gives the counit laws. The antipode isFor example $(TS(T){)}_{11}=ad-q^{-1}bc=1$ and $(S(T)T{)}_{11}=da-qbc=1$, while the off-diagonal entries vanish by the commutation relations. Thus this is a Hopf algebra, not merely an algebra presentation.

Take the right coaction of the coordinate Hopf algebra of quantum SL2 on the quantum plane to beIt is extended multiplicatively and sends $1$ to $1\otimes 1$. The matrix comultiplication and counit verify the coaction identities on $x,y$. To check that it descends to the quantum plane, expandThe last bracket vanishes because $bc=cb$ and $da-ad=(q-q^{-1})bc$.

For the requested calculation, $ca=qac$ and $yx=qxy$ giveMultiply by $\delta (y)$, put all monomials into the order $x^r{y}^s$, and use $bc=cb$. The answer isA left-coaction convention transposes where the matrix coefficients occur; the displayed answer fixes the right coaction throughout.

For an invertible braiding $C$ satisfying the braid equation on $V\otimes V$, begin with the free associative algebra over a field on matrix coefficients $t_{ij}$, with $\Delta t_{ij}={\sum }_k{t}_{ik}\otimes t_{kj}$ and $\epsilon (t_{ij})={\delta }_{ij}$. Impose precisely the coefficients ofThis quotient is the FRT bialgebra, or Faddeev-Reshetikhin-Takhtajan construction. The relations form a coideal because composing two coactions that preserve $C$ again preserves $C$; the trivial coaction also preserves it, so the counit descends. Universality follows because any such coaction must satisfy exactly these coefficient relations.

For the given $C$, take $t_{11}=a$, $t_{12}=b$, $t_{21}=c$, $t_{22}=d$. Thus $\delta (e_1)=e_1\otimes a+e_2\otimes c$ and $\delta (e_2)=e_1\otimes b+e_2\otimes d$. The coefficient relations, after cancellation of $q^{-1/2}$, areThese are exactly the defining relations of $M_q(2)$ in the inverse-parameter convention of Question 1, without setting the quantum determinant equal to one. They therefore define mutually inverse algebra homomorphisms: each named generator maps to the corresponding $t_{ij}$, and conversely. The matrix comultiplication and counit are the same, so the isomorphism is also one of bialgebras.

Here is the requested check of $ba=qab$ directly from $C$. Apply colinearity to $e_1\otimes e_2$ and compare the coefficient of $e_1\otimes e_1$. On the left, $C(e_1\otimes e_2)=q^{-1/2}{e}_2\otimes e_1$, giving $q^{-1/2}ba$. On the right, the corresponding term of $\delta (e_1\otimes e_2)$ is $e_1\otimes e_1\otimes ab$, and $C(e_1\otimes e_1)=q^{1/2}{e}_1\otimes e_1$. Hence $q^{-1/2}ba=q^{1/2}ab$, proving the relation. The other coefficient comparisons give the remaining displayed relations; no additional determinant relation is imposed in an FRT bialgebra.

The printed input $x^r{x}^s$ is a genuine typo in the PDF as well as in the TeX: $E$ annihilates every power of $x$, whereas the displayed right-hand side is generally nonzero. The intended input is $x^r{y}^s$, and the calculation below proves that corrected identity.

The given comultiplications make $C_q$ a coalgebra, with counit one on $K,K^{-1},I$ and zero on $E,F$. Write $I_0=p(I)$, $K_0=p(K)$ and $E_0=p(E)$, and similarly for $F$. The measuring coalgebra identities areThe group-like elements act multiplicatively. Although the statement omits an explicit specification of $p(I)$, its identity action follows from the other data and the quantum plane relation. Applying $E_0$ to $yx=qxy$ gives $q{x}^2=q{I}_0(x)x$, so cancellation in the noncommutative domain gives $I_0(x)=x$. Applying $F_0$ to the same relation gives $q{y}^2=qy{I}_0(y)$, so $I_0(y)=y$. Thus $I_0$ is the identity on the entire quantum plane.

Now $E_0(x^r)=0$ by induction. Suppose $E_0(y^{s-1})=[s-1]x{y}^{s-2}$. The measuring rule and $y^{s-1}x=q^{s-1}x{y}^{s-1}$ giveThe induction starts with $E_0(y)=x$ and $[1]=1$. A final product rule givesFor $s=0$ the answer is zero. By contrast the literal printed input has $\boxed{p(E)(x^r{x}^s)=0}$; it cannot satisfy the printed right-hand side for $s=1$.

We use right comodules and write their coactions as $u\mapsto \sum u_{(0)}\otimes u_{(1)}$. A coquasitriangular structure gives the braiding on these comodulesIn an arbitrary symmetric monoidal category this notation abbreviates a composite of the two coactions, the ambient symmetry, and $\gamma$; it does not assume that the objects have elements. The coquasitriangular axioms ensure that this composite is a comodule morphism, is invertible using the convolution inverse of $\gamma$, and satisfies the two hexagon laws. More explicitly, in scalar notation those laws come fromwhile the comodule-morphism condition isUnit normalizations give the unit constraints. These descriptions are identities of morphisms in the ambient symmetric category, with the displayed reordering carried by its symmetry.

Apply the preceding self-braiding result to the regular right comodule $(H,\Delta )$. Its Yang–Baxter operator isCompose its braid equation with the three counits. Expanding $R_{23}{R}_{12}{R}_{23}$ and cancelling the leading counit factors givesExpanding $R_{12}{R}_{23}{R}_{12}$ instead givesThey are equal by the Yang–Baxter operator equation. This is the required scalar identity:Indeed, after ${\Delta }^{\otimes 3}$ the ordered factors are $a_{(1)},a_{(2)},b_{(1)},b_{(2)},c_{(1)},c_{(2)}$. The two ambient symmetries on the left produce the three pairings $(a_{(1)},b_{(1)})$, $(a_{(2)},c_{(1)})$, $(b_{(2)},c_{(2)})$. On the right, the central symmetry produces $a_{(1)},a_{(2)},b_{(2)},b_{(1)},c_{(1)},c_{(2)}$, yielding exactly the other three pairings. Hence the calculation identifies the actual morphisms requested, also when the category is not a category of vector spaces.

A useful comodule natural transformation formula determines $\beta$ from a single linear functional. Defineusing the regular right comodule $H$. For any vector space $V$, the cofree right comodule $V\otimes H$ has coaction $1_V\otimes \Delta$. Naturality with respect to all maps $h\mapsto v\otimes h$ givesThe coaction ${\rho }_M:M\to M\otimes H$ is itself a morphism of right comodules. Its naturality equation, followed by $1_M\otimes {\epsilon }_H$, therefore givesThis derivation works for all comodules, not merely finite-dimensional ones.

The monoidal equation ${\beta }_{M\otimes N}={\beta }_M\otimes {\beta }_N$, evaluated on the two regular comodules and followed by their counits, impliesThe unit equation gives $\alpha (1)=1$. Thus $\alpha$ is a unital algebra homomorphism over a field $H\to k$. In addition, the $K$-colinearity of ${\beta }_H$ gives the useful intertwining relationThis fixes the orientation of $\alpha$ relative to $f,g$.

For the convolution product for coalgebra maps, define ${\alpha }^{-1}=\alpha S$. Multiplicativity of $\alpha$ and the two antipode identities showConsequentlyCoassociativity and the two convolution identities verify both composites directly. Since ${\beta }_M$ was a $K$-comodule morphism, its linear inverse is also a $K$-comodule morphism. Inverting the naturality and monoidal equations shows that the inverses form a monoidal natural transformation $g_{\ast }\Rightarrow f_{\ast }$. Every such monoidal transformation is therefore invertible, without requiring a bijective antipode.

Use right comodules with coaction $u\mapsto \sum u_{(0)}\otimes u_{(1)}$. Fix the convention that $\diamond$ is induced by $m$ and $\bullet$ by $n$. We classify lax monoidal structures first; the strong case is identified at the end.

Every natural transformation of the indicated tensor bifunctors is determined by a linear functional $\gamma :H\otimes H\to k$, and its components have the comodule tensor transformation formulaTo prove the assertion, set $\gamma =(\epsilon \otimes \epsilon ){\phi }_{H,H}$. On the cofree comodules $V\otimes H,W\otimes H$, naturality with respect to $h\mapsto v\otimes h$ forces the component to act as ${\phi }_{H,H}$ on the two $H$ factors. Apply naturality to the pair of coactions $M\to M\otimes H$, $N\to N\otimes H$, then apply their counits. This yields the formula. Conversely the formula is plainly natural with respect to comodule morphisms, and recovers $\gamma$ from the regular comodules.

It remains to impose exactly three kinds of equations. First, the component must be $H$-colinear from the tensor with product $n$ to the tensor with product $m$. Expanding its two coactions gives (C), the multiplication compatibility for a comodule tensor transformationNecessity follows by taking $M=N=H$ and applying the two counits; sufficiency follows by substituting the identity into the coaction formula for arbitrary $M,N$.

Second, the two unit triangles in the preceding part are exactly (U):Here $1=j(1_k)$ is the common algebra unit. These identities are forced by the regular comodule and sufficient by the counit law.

Third, expand the associativity diagram on three right comodules. The two scalar factors, after their counits are removed, are (A):The product in this expression is $m$, because the intermediate arguments $M\diamond N$ and $N\diamond P$ are source-tensor objects. Necessity follows from the three regular comodules; sufficiency is the same substitution for arbitrary coactions. Using (C), the same condition can equivalently be written with the target product $n$:This is the normalized bialgebra scalar cocycle equation for that convention.

Thus the required structures correspond bijectively to all linear maps $\gamma$ satisfying (C), (U) and (A). No convolution invertibility has been imposed for a lax monoidal functor structure. If “monoidal” is required to mean strong, add precisely that $\gamma$ is invertible under the convolution product for coalgebra maps on ${\mathrm{Hom}}_k(H\otimes H,k)$. Its convolution inverse $\bar{\gamma }$ supplies the inverse component by the same displayed formula; conversely natural inverse components recover a convolution inverse by the regular-comodule argument. In that case (C) is the explicit bialgebra cocycle twist relationThis also records which multiplication lies on each side of the twist. Interchanging the names of $m,n$ interchanges the two tensor conventions, rather than silently changing the equation.

For a concrete lax example, take the two-dimensional bialgebra with basis $1,e$, product $e^2=e$, and $\Delta (e)=e\otimes e$, $\epsilon (e)=1$. Set $m=n$ and $\gamma (1,1)=\gamma (1,e)=\gamma (e,1)=1$, $\gamma (e,e)=0$. Normalization and (C) hold, and (A) follows by checking the eight basis triples: triples involving $1$ reduce to normalization, and the triple $(e,e,e)$ has both sides zero. But the $(e,e)$ tensor component is zero, so this lax monoidal functor is not strong.

Over a field with $q\ne 0$ and $q^2\ne 1$, this Hopf algebra has generators $E,F,K,K^{-1}$ and relations $KE=q^{2}EK$, $KF=q^{-2}FK$ and $EF-FE=(K-K^{-1})/(q-q^{-1})$. One comultiplication convention is $\Delta E=E\otimes K+1\otimes E$, $\Delta F=F\otimes 1+K^{-1}\otimes F$ and $\Delta K=K\otimes K$. Its counit takes $E,F$ to zero and $K$ to one. The antipode takes $E$ to $-E{K}^{-1}$, $F$ to $-KF$ and $K$ to $K^{-1}$.