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A comonoid morphism $f:H\to K$ induces a functor $f_{\ast }$ on right comodules, replacing $\rho$ by $(1\otimes f)\rho$ and preserving underlying objects and arrows. For bimonoids, it is strict monoidal exactly when $f$ is also a monoid morphism.

For the given bimonoids, use right comodules. The corestriction functor for comodules associated to a comonoid morphism $f:H\to K$ keeps underlying objects and morphisms, and replaces a coaction ${\rho }_M$ by $(1\otimes f){\rho }_M$. The comonoid-morphism axioms ensure that this is a $K$-coaction.

The tensor coaction for two $H$-comodules in the ambient braided monoidal category isand the unit coaction is $j_H:I\to H$. If $f$ is also a monoid morphism, then $f{m}_H=m_K(f\otimes f)$ and $f{j}_H=j_K$. Substituting these identities, and using naturality of the ambient braiding, shows that corestriction preserves both tensor and unit coactions exactly. Its structural maps are identities, so it is a strict monoidal functor.

Conversely, suppose this induced functor is strict monoidal. Apply equality of the tensor coactions to the two regular right comodules $(H,{\Delta }_H)$. Then apply ${\epsilon }_H\otimes {\epsilon }_H$ to their two underlying $H$ factors. The counit laws and naturality of the braiding remove those factors and leaveEquality on the unit comodule similarly gives $f{j}_H=j_K$. Thus $f$ is a monoid morphism. The criterion is exactlyThe regular-comodule argument uses only the counit laws; it requires no elementwise or finite-dimensional assumption on the ambient category.