Mistake in Wikipedia Entry "Coalgebra" - MathOverflow

Mistake in Wikipedia Entry "Coalgebra"

2012-06-01T16:00:22Z

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?

Answer by Ralph for Mistake in Wikipedia Entry "Coalgebra"

2012-06-01T16:33:34Z

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

Answer by Gjergji Zaimi for Mistake in Wikipedia Entry "Coalgebra"

2012-06-01T17:16:34Z

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.

Answer by Tilman for Mistake in Wikipedia Entry "Coalgebra"

2013-02-19T10:39:47Z

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 $4f_1 = \Delta(f_2)$ to $im(f) \otimes im(f)$ would have to have order $4$, but $\overline e_2$ and $f_2$ only have order $2$.