Let $C$ be a coalgebra and $J$ a subvector space of $C$. How to show that $J$ is a coideal of $C$ if and only if $J^{\perp}$ is a subalgebra of $C^*$? Here $J^{\perp} = \{f\in C^*: \langle f, v \rangle = 0, \forall v \in J\}$.
My solution: suppose that $J$ is a coideal of $C$. Then we have $\Delta(J) \subset J \otimes C + C \otimes J$. We need to show that $J^{\perp}$ is closed under multiplication. Let $f^*, g^* \in J^{\perp}$. Then $\langle f^*, v\rangle = \langle J^*, v\rangle = 0$, for all $v \in J$. We have $$ \langle f^*g^*, v\rangle = \langle f^*, v_{(1)}\rangle \langle g^*, v_{(2)}\rangle. $$ Since $\Delta(J) \subset J \otimes C + C \otimes J$, we have $$ \langle f^*g^*, v\rangle = \langle f^*, v_{(1)}\rangle \langle g^*, v_{(2)}\rangle = 0. $$ Therefore $f^*g^* \in J^{\perp}$.
How to prove the other direction? Thank you very much.
If $J$ is not a coideal, then $\Delta(v)=u\otimes w+(\mbox{other terms})$ where $u,w\notin J$. Let $f^*$ (resp. $g^*$) be the coordinate function on $u$ (resp. $w$). Then $$\langle f^*g^*,v\rangle=\langle f^*,u\rangle\langle g^*,w\rangle=1.$$