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Theo already proved in the comments that the image of a coalgebra map is always a subcoalgebra of the codomain. Here is an example where the kernel is not a coideal, taken from Nichols and Sweedler's "Hopf Algebras and Combinatorics" (also exercise 2.15.5 in "Corings and Comodules" by Brzeziński and Wisbauer):

Let $C_1=\mathbb Z\oplus \mathbb Z/2\mathbb Z\oplus\mathbb Z$ with $c_0=(1,0,0),c_1=(0,1,0),c_2=(0,0,1)$ and $$\Delta(c_0)=c_0\otimes c_0$$ $$\Delta(c_1)=c_0\otimes c_1+c_1\otimes c_0$$ $$\Delta(c_2)=c_0\otimes c_2+c_1\otimes c_1+c_2\otimes c_0$$ Let $C_2=\mathbb Z\oplus \mathbb Z/4\mathbb Z$ with $d_0=(1,0),d_1=(0,1)$ and $$\Delta(d_0)=d_0\otimes d_0$$ $$\Delta(d_1)=d_0\otimes d_1+d_1\otimes d_0$$ Now take the coalgebra map $f: C_1\to C_2$ that sends $$c_0\to d_0,c_1\to 2d_1,c_2\to 0,$$ its kernel is $c_2\mathbb Z$and the kernel of $f\otimes f$ is $c_2\otimes C_1+C_1\otimes c_2$. However $\Delta(c_2)\in c_2\in \operatorname{ker}(f)$ but $\Delta(c_2)\notin c_2\otimes C_1+C_1\otimes c_2$ so the kernel of $f$ is not a coideal.

If instead we restrict to $f':C_1\to f(C_1)$, we no longer have $\Delta(c_2)\in \operatorname{ker}(f')$. And in fact $\operatorname{ker}(f')$ is a coideal of $C_1$.

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Theo already proved in the comments that the image of a coalgebra map is always a subcoalgebra of the codomain. Here is an example where the kernel is not a coideal, taken from Nichols and Sweedler's "Hopf Algebras and Combinatorics" (also exercise 2.15.5 in "Corings and Comodules" by Brzeziński and Wisbauer):

Let $C_1=\mathbb Z\oplus \mathbb Z/2\mathbb Z\oplus\mathbb Z$ with $c_0=(1,0,0),c_1=(0,1,0),c_2=(0,0,1)$ and $$\Delta(c_0)=c_0\otimes c_0$$ $$\Delta(c_1)=c_0\otimes c_1+c_1\otimes c_0$$ $$\Delta(c_2)=c_0\otimes c_2+c_1\otimes c_1+c_2\otimes c_0$$ Let $C_2=\mathbb Z\oplus \mathbb Z/4\mathbb Z$ with $d_0=(1,0),d_1=(0,1)$ and $$\Delta(d_0)=d_0\otimes d_0$$ $$\Delta(d_1)=d_0\otimes d_1+d_1\otimes d_0$$ Now take the coalgebra map $f: C_1\to C_2$ that sends $$c_0\to d_0,c_1\to 2d_1,c_2\to 0,$$ its kernel is $c_2\mathbb Z$ and the kernel of $f\otimes f$ is $c_2\otimes C_1+C_1\otimes c_2$. However $\Delta(c_2)\in \operatorname{ker}(f)$ but $\Delta(c_2)\notin c_2\otimes C_1+C_1\otimes c_2$ so it's the kernel of $f$ is not a coideal.

If instead we restrict to $f':C_1\to f(C_1)$, we no longer have $\Delta(c_2)\in \operatorname{ker}(f')$. And in fact $\operatorname{ker}(f')$ is a coideal of $C_1$.

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Theo already proved in the comments that the image of a coalgebra map is always a subcoalgebra of the codomain. Here is an example where the kernel is not a coideal, taken from Nichols and Sweedler's "Hopf Algebras and Combinatorics" (also exercise 2.15.5 in "Corings and Comodules" by Brzeziński and Wisbauer):

Let $C_1=\mathbb Z\oplus \mathbb Z/2\mathbb Z\oplus\mathbb Z$ with $c_0=(1,0,0),c_1=(0,1,0),c_2=(0,0,1)$ and $$\Delta(c_0)=c_0\otimes c_0$$ $$\Delta(c_1)=c_0\otimes c_1+c_1\otimes c_0$$ $$\Delta(c_2)=c_0\otimes c_2+c_1\otimes c_1+c_2\otimes c_0$$ Let $C_2=\mathbb Z\oplus \mathbb Z/4\mathbb Z$ with $d_0=(1,0),d_1=(0,1)$ and $$\Delta(d_0)=d_0\otimes d_0$$ $$\Delta(d_1)=d_0\otimes d_1+d_1\otimes d_0$$ Now take the coalgebra map $C_1\to C_2$ that sends $$c_0\to d_0,c_1\to 2d_1,c_2\to 0,$$ its kernel is $c_2\mathbb Z$ and $\Delta(c_2)\notin c_2\otimes C_1+C_1\otimes c_2$ so it's not a coideal.