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I want to show, that if $I$ is a coideal in a coalgebra $A$, then $A/I$ is a coalgebra. To be a coideal means $\Delta(I)\subset A\otimes I+I\otimes A$ and $\varepsilon(I)=0$. We can use the induced maps from $A$, which means $$\Delta_{A/I}(\overline{a})=\Delta(a)$$ $$\varepsilon_{A/I}(\overline{a})=\varepsilon(a)$$ Here the $\overline{a}$ denote the class of $a$ in the quotient $A/I$. That $\varepsilon_{A/I}$ is well-defined is easy, but I don't see why $\Delta_{A/I}$ should be.

My idea: Suppose $a\equiv b$. Then $b-a\in I$. We have to show that $\Delta(a)=\Delta(b)$, but I don't see how to use that $\Delta(I)\subset A\otimes I+I\otimes A$. Can someone help me? Maybe I get the wrong definition?

Thanks a lot.

Egbert53
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The sum $I\otimes A + A\otimes I$ is contained in the kernel of $A\otimes A\to A/I \otimes A/I$ (and, if you are working over a field, is actually the whole kernel). You can then finish your proof as follows: You have $a-b\in I$, so $\Delta(a) - \Delta(b)=\Delta(a-b)\in I\otimes A + A\otimes I$ by assumption. Hence $\Delta(a) = \Delta(b)$ in $A/I\otimes A/I$ as desired.

Hanno
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    Over any commutative ring, the kernel of $A \otimes A \to A/I \otimes A/I$ is the image of $I \otimes A ~\oplus ~ A \otimes I \to A \otimes A$ (since $\otimes$ is right exact). The usual definition of a coideal ( $\Delta(I) \subseteq I \otimes A+A \otimes I$) doesn't even make sense since $I \otimes A \to A \otimes A$ doesn't have to be injective. I think that the correct definition is $\Delta(I) \subseteq \overline{I \otimes A} + \overline{A \otimes I}$, where $\overline{I \otimes A}$ denotes the image of $I \otimes A \to A \otimes A$. Namely, then $A/I$ acquires the structure of a coalgebra. – Martin Brandenburg Jan 28 '16 at 15:44
  • Why does the kernel of $(\pi \otimes \pi) \circ \Delta$ coincide with $I \otimes C + C \otimes I$? Can't $\Delta$ itself have a non-trivial kernel? – Daigaku no Baku Feb 29 '24 at 10:37