The kernel of a morphism of co-rings is a co-ideal

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.

• What is $\pi$ here? – Alex Clark Apr 6 '18 at 14:03
• Sorry, $\pi$ is the natural projection of $C$ unto $C/\text{Ker}(\phi)$ – tomak Apr 6 '18 at 14:17
• Since you are speaking generally about corings (and not especially about coalgebras) could you please make explicit the definition of the notion of co-ideal you are using ? – KonKan Apr 8 '18 at 21:51

The statement:

The Kernel of a coring morphism $\phi:C\rightarrow C'$ between two $R$-coring's $(C,\Delta,\epsilon)$ and $(C',\Delta',\epsilon')$ is a coideal.

is not true in general:

• It is true if one considers the special case of coalgebras over a field $k$ (a proof of this, can be found at Hopf algebras: an introduction, Dascalescu et al, p. 25-26, proposition 1.4.9),
• However, in the general case that we have an $A$-coring $C$ instead of a coalgebra over a field $k$ (now the linear structure of $C$ is that of an $A$-bimodule ${}_AC_A$, where $A$ is an arbitrary -that is: non-commutative in general- algebra) the statement is not valid as it is: In this case, the coideal is by definition the kernel of a surjective $A$-coring morphism (and not of any $A$-coring morphism). For this last case, see the proof and futher details at: Corings and Comodules, Brzezinski et al, proposition 17.14, p. 177-178.
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)
• Thanks for the references. I am taking my statement from a paper by Sweedler "The prequel to the Jacobson-Bourbaki theorem" where it is clearly stated that the kernel of a coring morphism is a coideal. I'll put the definitions as an edit – tomak Apr 10 '18 at 8:14
• @tomak: thank you for your edit. It contributes to clarifying the problem. However, i tried a little search on the Sweedler's paper you are citing and i was not able to find it online. Could you add some link to it? – KonKan Apr 10 '18 at 20:03
• I got the pdf from my teacher. Any way I could send it to you? – tomak Apr 11 '18 at 9:00
• Could you possibly use an upload-a-file-to-share facility, such as for example: expirebox.com and share the corresponding link here? – KonKan Apr 11 '18 at 15:27
• @tomak: you could also send me an email. I d' be glad to take a look at the paper you are citing. Of course you should also ask your teacher ;) – KonKan Apr 16 '18 at 0:52