Mistake in Wikipedia Entry "Coalgebra" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:29:54Z http://mathoverflow.net/feeds/question/98581 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98581/mistake-in-wikipedia-entry-coalgebra Mistake in Wikipedia Entry "Coalgebra" Ago Szekeres 2012-06-01T16:00:22Z 2013-02-19T10:39:47Z <p>Consider the following quote from the Wikipedia entry <em>Coalgebra</em>:</p> <blockquote> <p>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 $C_2$.</p> </blockquote> <p>I can't see any qualifiers preceding or succeeding the statement. Am I missing something obvious here, or is this just plain wrong?</p> <p>Do there not exists kernels of coalgebra maps that are not coideals?</p> http://mathoverflow.net/questions/98581/mistake-in-wikipedia-entry-coalgebra/98585#98585 Answer by Ralph for Mistake in Wikipedia Entry "Coalgebra" Ralph 2012-06-01T16:33:34Z 2012-06-01T22:30:33Z <p>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": </p> <p>Prop. 1.4.4: The image of a coalgebra morphism is a subcoalgebra </p> <p>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) </p> <p><strong>Added:</strong> 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. </p> <p>For instance, in Gjergji's example $f$ isn't surjective since $\mathbb{Z}/4 \nsubseteq \operatorname{im}(f)$. </p> http://mathoverflow.net/questions/98581/mistake-in-wikipedia-entry-coalgebra/98589#98589 Answer by Gjergji Zaimi for Mistake in Wikipedia Entry "Coalgebra" Gjergji Zaimi 2012-06-01T17:16:34Z 2012-06-02T00:12:01Z <p>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 BrzeziĆski and Wisbauer): </p> <p>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.</p> http://mathoverflow.net/questions/98581/mistake-in-wikipedia-entry-coalgebra/122284#122284 Answer by Tilman for Mistake in Wikipedia Entry "Coalgebra" Tilman 2013-02-19T10:39:47Z 2013-02-19T10:39:47Z <p>Here's a counterexample to show that the image of a coalgebra map isn't always a coalgebra (adapted from the Nichols-Sweedler counterexamples).</p> <p>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.</p> <p>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 $2$.</p>