I would like to show that
The Kernel of a coring morphism $\phi:C\rightarrow C'$ between two $R$-corings $(C,\Delta,\epsilon)$ and $(C',\Delta',\epsilon')$ is a coideal.
The only point that I can't show is that $$\Delta(\text{Ker}(\phi))\subset \text{Ker}(\pi\otimes\pi)$$
This should be straightforward but I can't show it.
EDIT:
$\pi$ is the natural projection of $C$ unto $C/\text{Ker}(\phi)$
The definition I use are
A coring morphism from an $R$-coring $(C,\Delta,\epsilon)$ to another $R$-coring $(C',\Delta',\epsilon')$ is a $R$-bimodule morphism $\phi$ such that $\epsilon'\circ\phi = \epsilon$ and $(\phi\otimes\phi)\circ\Delta = \Delta'\circ\phi$.
A coideal $I$ of the $R$-coring $C$ is a $R$-subbimodule of $C$ such that $I\subset \text{Ker}(\epsilon)$ and \begin{align*} \Delta(I) \subset \text{Ker}(\pi\otimes\pi) \end{align*} where $\pi:C\rightarrow C/I$ is the cannonical projection map.
The statement:
is not true in general:
The same situation holds in the case of coalgebras over a commutative ring $R$ (for a proof of the case where the "scalars" are elements of a commutative ring $R$, see: Corings and Comodules, Brzezinski et al., proposition 2.4, p. 9-10)