Consider the following quote from the Wikipedia entry *Coalgebra*:



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> The kernel of every coalgebra morphism $f : C\_1 \to C\_2$ is a coideal in $C\_1$, and the image is a subcoalgebra of $C\_2$.
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> 
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I can't see any qualifiers preceding or succeeding the statement. Am I missing something obvious here, or is this just plain wrong?


Do there not exists kernels of coalgebra maps that are not coideals?



Mistake in Wikipedia Entry "Coalgebra"
As the statement that a kernel is a coideal isn't treated in the comments let me give the following reference from Sweedler's book "Hopf Algebras": 


Prop. 1.4.4: The image of a coalgebra morphism is a subcoalgebra 


Theorem 1.4.7 b): If it's an coalgebra over a field, then the kernel of a coalgebra morphism is a coideal. (compare Gjergji's counterexample in the general case) 


**Added:** An inspection of Sweedler's proof of Th. 1.4.7 b) shows that the crucial property is $\ker(f \otimes f) = \ker f \otimes C\_1 + C\_1 \otimes \ker f$. This always holds over a field but is usually false over a comm. ring. However, if $f$ is surjective this identity holds over any ring. In particular, over any comm. ground ring, the kernel of a surjective coalgebra morphism is a coideal. 


For instance, in Gjergji's example $f$ isn't surjective since $\mathbb{Z}/4 \nsubseteq \operatorname{im}(f)$. 


