Definition morphism of Hopf algebra

Question

this is my first post, so let me know if you need some more information.

I am currently studying Hopf Algebras and and exercise I have tells me to show that $f:H \rightarrow H'$ is a morphism of Hopf Algebras, given that $H, H'$ are bialgebras and $f$ is a morphism of bialgebras.

My question is, what exactly is a morphism of Hopf algebras. I can't seem to find a precise definition online. Some sources say it preserves this antipode, but I can't get my head around what this actually means.

Thanks

Answer 1

Usually the definition is as follows, see Definition $4.2.4$ in the book Hopf Algebras: An Introduction ( I found this online):

Let $H$ and $B$ two Hopf algebras. A map $f\colon H\rightarrow B$ is called a morphism of Hopf algebras if it is a morphism of bialgebras.

Answer 2

Yes, one can define a morphism $f:H\rightarrow H'$ of bialgebras $H$ and $H'$ over a field $k$ as a (unital) algebra and a (counital) coalgebra map. If $H$ has an antipode $S_H$ and $H'$ has an antipode $S_{H'}$, then you could say that a bialgebra morphism is a Hopf algebra morphism if it also satisfies $f\circ S_H = S_{H'}\circ f$. However it happens that this follows automatically for any bialgebra map. For the proof (see for example Schneider's lecture notes) you can argue using the uniqueness of inverses in the convolution algebra $\operatorname{Hom}(H,H')$. If $\epsilon_H:H\rightarrow k$ is the counit of $H$ and $\eta_{H'}:k\rightarrow H'$, with $\eta_{H'}(\lambda) = \lambda 1_{H'}$ is the unit of $H'$, then the identity element of $\operatorname{Hom}(H,H')$ is the map $\eta_{H'}\circ\epsilon_H$. Now the "trick" is to show that $f\circ S_H$ and $S_{H'}\circ f$ are both inverses of $f$ in $\operatorname{Hom}(H,H')$ (w.r.t. convolution product). So for all $h\in H$: $$((f\circ S_H) * f)(h) = \sum_{(h)} f(S(h_1))f(h_2) = f\left(\sum_{(h)} S(h_1)h_2\right) = \epsilon_H(h) f(1_H) = \epsilon_H(h)1_{H'}$$ using that $f$ is an unital algebra homomorphism. Hence $(f\circ S_H)*f=\eta_{H'}\circ \epsilon_H$ shows that $f\circ S_H$ is a left inverse of $f$. On the other hand for all $h\in H$: $$(f* (S_{H'}\circ f))(h) = \sum_{(h)} f(h_1)S_{H'}(f(h_2)) = \sum_{(f(h))} f(h)_1 S_{H'}(f(h)_2) = \epsilon_{H'}(f(h)) = \epsilon_H(h)1_{H'}$$ using that $f$ is a counital coalgebra homomorphism. Hence $f*(S_{H'}\circ f)=\eta_{H'}\circ\epsilon_H$ shows that $S_{H'}\circ f$ is a right inverse of $f$. By the uniqueness of inverses, $f\circ S_H = S_{H'}\circ f$.