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 it also this identity holds over any ring. In particular, over any comm. ground ring, the kernel of a surjective coalgebra morphism is a coideal.
For exampleinstance, in Gjergji's example $f$ isn't surjective since $\mathbb{Z}/4 \nsubseteq \operatorname{im}(f)$.
