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For a transformation from the tensor built using $n$ to the tensor built using $m$, colinearity is $m\ast \gamma =\gamma \ast n$: $\sum m(a_{(1)},b_{(1)})\gamma (a_{(2)},b_{(2)})=\sum \gamma (a_{(1)},b_{(1)})n(a_{(2)},b_{(2)})$. Scalar maps enter this convolution product for coalgebra maps via the common algebra unit.

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.