Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 122 / 5

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.

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.

The induced internal hom for modules over a Hopf algebra on ${\mathrm{Hom}}_k(H,H)$ iswhere both occurrences of $H$ have their left regular action. If $T$ is a morphism of left $H$-modules, thenSo every such morphism is an invariant vector of a module over a Hopf algebra.

For the converse, a calculation valid for every $k$-linear $T$ isIf $T$ is invariant, its left-hand side is instead $\sum \epsilon (h_{(1)})T(h_{(2)}x)=T(hx)$. Hence $T(hx)=hT(x)$ for all $h,x$, which is precisely the R-module homomorphism condition.

In fact an endomorphism of the left regular module is determined by $a=T(1)$ and has $T(x)=xa$. Conversely each right multiplication map is left $H$-linear. Thus the invariant space has the explicit descriptionAs a vector space it is isomorphic to $H$, and under composition its algebra is the opposite algebra $H^{\mathrm{op}}$. No dimension or semisimplicity hypothesis is needed.