Context: Let $\phi:A\to B$ be a morphism of Hopf algebras. To show that $\ker\phi$ is a Hopf ideal of $A$, we must verify that $\Delta(\ker\phi)\subseteq \ker\phi\otimes A+A\otimes\ker\phi$, where $\Delta:A\to A\otimes A$ is the co-multiplication.
I have showed that $\Delta(\ker\phi)\subseteq \ker(\phi\otimes\phi)$ using the compatibility conditions for morphism of Hopf algebras.
It turns out that showing $\ker(\phi\otimes\phi)\subseteq \ker\phi\otimes A+A\otimes\ker\phi$ is an easy exercise in linear algebra, but I can't really find a satisfactory proof.
My attempt: Let $(v_i)_{i\in I}$ be a basis of $\ker(\phi)$ and extend it to a basis $(v_i)_{i\in J}$ of $A$. Then any element $x$ in $A\otimes A$ is of the form $$ x=\sum_{i,j\in J}x_{ij}v_i\otimes v_j.$$
Then $x\in \ker(\phi\otimes\phi)\iff \sum_{i,j\in J}x_{ij}\phi(v_i)\otimes \phi(v_j)=0$. Now, I know for sure that $(\phi(v_i))_{i\in J}$ is a linear independent set in $B$.
QUESTION 1: Is the same true for $(\phi(v_i)\otimes \phi(v_j))_{i,j\in J}$ in $B\otimes B$? If so, this would imply $x_{ij}=0,\forall i,j$.
I believe the final reasoning is then as follows: by linearity of $\phi$ and properties of the tensor product, we can put some $x_{ij}$ with $v_i$, creating terms of the form $\phi(x_{ij}v_i)\otimes \phi(v_j)\in\ker\phi\otimes A$, but we could also put them with $v_j$, creating terms of the form $\phi(v_i)\otimes \phi(x_{ij}v_j)\in A\otimes\ker\phi$.
QUESTION 2: This kind of seems like a little bit of cheating, but is it ok?