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Consider the following quote from the Wikipedia entry Coalgebra:

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 $C_2$.

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?

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There are lots of mistakes on Wikipedia. – Spiro Karigiannis Jun 1 '12 at 16:27
Certainly if $f$ is a morphism of coalgebras, then $\Delta_{C_2}f(c_1) = (f\otimes f)(\Delta_{C_1}c_1) \in f(C_1) \otimes f(C_1)$, so that the image is a subcoalgebra. Or am I confused? The dual statement for algebras is that the coimage of an algebra morphism $A_1 \to A_2$ is a quotient algebra of $A_1$, which is also certainly true. Incidentally, I've never liked the word "coideal" for this notion — a coideal should be a cokernel of a map. – Theo Johnson-Freyd Jun 1 '12 at 16:32
Coalgebra morphisms are linear (i.e. vector space) morphisms that respect the comultiplication, so the image of a coalgebra is also a coalgebra. They also send the counit to the counit, so the image is a subcoalgebra of the codomain, once you restrict the counit. – Zack Wolske Jun 1 '12 at 16:36
If you see a mathematical mistake on Wikipedia, it immediately becomes your fault if you don't click "edit" and fix it. – Brendan McKay Jun 3 '12 at 14:12
I think that "coimage" may be an unfortunate coinage. – Tom Goodwillie May 25 '14 at 17:04
up vote 8 down vote accepted

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 $\mathbb{Z}/4 \nsubseteq \operatorname{im}(f)$.

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In Sweedler's book, one has a ground field, while in Gjergji's answer the coalgebra is a 'coring', really. – Mariano Suárez-Alvarez Jun 1 '12 at 17:21
Yes, I should had quoted it properly. Thanks for the hint. – Ralph Jun 1 '12 at 17:32
@Mariano: Thanks for making the distinction between the two results. The Nichols - Sweedler example was the one that inspired my question. I got confused by field/coring choice. – Ago Szekeres Jun 1 '12 at 18:00
Regarding your addendum: since the image of $f$ is a subcoalgebra, restricting the codomain of $f$ to it gives a surjective map with the same kernel. Something breaks, no? – Mariano Suárez-Alvarez Jun 1 '12 at 20:27
So that means that while the kernel of $f:C_1\to C_2$ is not a coideal of $C_1$, the kernel of $f':C_1\to f(C_1)$ is. – Mariano Suárez-Alvarez Jun 1 '12 at 22:08

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 $c_2\mathbb Z$. However $c_2\in \operatorname{ker}(f)$ but $\Delta(c_2)\notin c_2\otimes C_1+C_1\otimes c_2$ so the kernel of $f$ is not a coideal.

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What is the kernel of $f'$? I am surely messing up my calculation... :( – Mariano Suárez-Alvarez Jun 1 '12 at 22:45
I think $\Delta(c_2) \in \ker(f)$ should be $c_2 \in \ker(f)$. Also note that $\ker(f') = \ker(f)$. – Ralph Jun 1 '12 at 23:41
Thanks guys, I got confused on the calculations there for a moment. – Gjergji Zaimi Jun 2 '12 at 0:14

Here's a counterexample to show that the image of a coalgebra map isn't always a coalgebra (adapted from the Nichols-Sweedler counterexamples).

Let $C = \mathbb{Z} \oplus \mathbb{Z}$ with $\Delta(e_1) = 0$, $\Delta(e_2)=e_1 \otimes e_1$ and $D = \mathbb{Z}/8 \oplus \mathbb{Z}/2$ with $\Delta(f_1) = 0$, $\Delta(f_2) = 4f_1 \otimes f_1$. (Here $e_i$ and $f_i$ denote the standard generators). Define a map $f\colon C \to D$ by $f(e_1) = 2f_1$ and $f(e_2) = f_2$. That's a coalgebra map.

Now the image of $f$ is $\mathbb{Z}/4 \oplus \mathbb{Z}/2$, I'm calling the generators $\overline e_1$ and $\overline e_2$. But it's not a sub-coalgebra. Any lift of $4f_1 \otimes f_1 = \Delta(f_2)$ to $im(f) \otimes im(f)$ would have to have order $4$, but $\overline e_2$ and $f_2$ only have order $2$.

These are non-counital coalgebras, but they can easily be made counital by taking the direct sum with $\mathbb{Z}$ and interpreting the above-defined $\Delta$ map as reduced coproduct $\overline\Delta$.

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