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.
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.