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?



Bialgebra maps and Hopf algebra maps
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 $\phi\circ S\_H$).


