Let $H$ and $H'$ be two Hopf algebras, and let $\phi:H \to H'$ be an bialgebra map. Then is $\phi$ automatically a Hopf algebra map?
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$\begingroup$ For commutative or cocommutative Hopf algebras, the answer is yes. Presumably this can be proved by a diagram chase, but there is also a quick way. A cocommutative bialgebra is a monoid object in the category of cocommutative coalgebras, and the Hopf algebras are the group objects. So in the cocommutative case, what you want is a special case of the fact that in a category with finite products, a morphism of monoid objects between two group objects is actually a morphism of group objects. This follows from the usual statement in Set using the Yoneda embedding. $\endgroup$– Alexander BettsCommented Sep 2, 2021 at 22:43
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Yes it is.
It is actually a standard result, for any hopf algebra, that under the circumstances you are describing any bialgebra map $\phi$ commutes with the antipode, i.e. we have: $$S_{H'}\circ \phi=\phi\circ S_H.$$
(The simplest way to show this is to consider $\operatorname{Hom}(H,H')$ as an algebra —under the convolution product— and to show that $\phi$ is convolution invertible with left inverse $S_{H'}\circ \phi$ and right inverse $\phi\circ S_H$).
answered Sep 2, 2021 at 22:46
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2$\begingroup$ . . . and then conclude the desired result from uniqueness of inverses? $\endgroup$ Commented Sep 2, 2021 at 23:30
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$\begingroup$ yes, that is right. $\endgroup$ Commented Sep 2, 2021 at 23:31
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$\begingroup$ you are welcome. And .. welcome to MO! $\endgroup$ Commented Sep 2, 2021 at 23:36