Timeline for Mistake in Wikipedia Entry "Coalgebra"
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 25, 2014 at 17:04 | comment | added | Tom Goodwillie | I think that "coimage" may be an unfortunate coinage. | |
| Feb 19, 2013 at 10:39 | answer | added | Tilman | timeline score: 5 | |
| Feb 18, 2013 at 16:41 | comment | added | Tilman | I suppose this thread counts as dead, but never mind. I don't understand Theo's argument that the image of a coalgebra is a coalgebra. How is your element not dependent on the choice of $c_1$? There may be no coherent lift of $\Delta$ on $im(f)$ to $im(f) \otimes im(f)$ unless $im(f) \otimes im(f) \to D \otimes D$ is injective. | |
| Jun 3, 2012 at 14:12 | comment | added | Brendan McKay | If you see a mathematical mistake on Wikipedia, it immediately becomes your fault if you don't click "edit" and fix it. | |
| Jun 1, 2012 at 20:30 | comment | added | Mariano Suárez-Álvarez | @Theo: half of the time, I think so too. But then a subcoalgebra of $C$ should be a quotient of $C$, and it becomes weird... | |
| Jun 1, 2012 at 19:39 | comment | added | darij grinberg | Okay, forget about the second way; it's probably an abuse of the word "subcoalgebra" even when $k$ is a field. | |
| Jun 1, 2012 at 19:36 | comment | added | darij grinberg | @Theo: I see at least three ways to define a "subcoalgebra" of a coalgebra $D$. The first is to define it as a $k$-submodule $M$ of $D$ satisfying $\Delta_D\left(M\right)\subseteq \left(\text{the image of the canonical map }M\otimes M\to D\otimes D\right)$. The second is to define it as a subobject in the category of coalgebras, i. e., a coalgebra $M$ with a coalgebra monomorphism $M\to D$. The third is to define it as a coalgebra $M$ with an injective coalgebra homomorphism (that's not the same as a monomorphism!) $M\to D$. Which of these are equivalent when $k$ is not a field? I don't know. | |
| Jun 1, 2012 at 18:00 | vote | accept | Ago Szekeres | ||
| Jun 1, 2012 at 17:16 | answer | added | Gjergji Zaimi | timeline score: 5 | |
| Jun 1, 2012 at 16:38 | history | edited | Ago Szekeres | CC BY-SA 3.0 |
deleted 87 characters in body
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| Jun 1, 2012 at 16:37 | comment | added | Ago Szekeres | @ Theo: Yes, you're right - stupid mistake on my part - I've edited accordingly. However, it still stands that the kernel of a coalgebra map need not be a coideal. | |
| Jun 1, 2012 at 16:36 | comment | added | Zack Wolske | 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. | |
| Jun 1, 2012 at 16:33 | answer | added | Ralph | timeline score: 8 | |
| Jun 1, 2012 at 16:32 | comment | added | Theo Johnson-Freyd | 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. | |
| Jun 1, 2012 at 16:27 | comment | added | Spiro Karigiannis | There are lots of mistakes on Wikipedia. | |
| Jun 1, 2012 at 16:00 | history | asked | Ago Szekeres | CC BY-SA 3.0 |