Let $(A,\Delta)=:F(G)$ be a **finite dimensional** $\mathrm{C}^*$-Hopf algebra, and so the algebra of functions on a *quantum group* $G$.

Let $J$ be a closed ideal in $F(G)$ and $\pi:F(G)\rightarrow F(G)/J$ the quotient map. Suppose that $J$ is such that
$$\Delta(J)\subset \ker(\pi\otimes\pi).$$
Such ideals are called *Woronowicz $\mathrm{C}^*$-ideals*.

When $A$ is commutative, so that $G$ is a finite group, Woronowicz $\mathrm{C}^*$-ideals correspond to functions that vanish on a distinguished subgroup $H$, and the quotient map, via $F(G)/J\cong F(H)$, $\pi_H$ given by: $$\pi_H\left(\sum_{t\in G}a_t\delta_t\right)=\sum_{t\in H}a_t\delta_t,$$ and $J=\{f\in F(G)\,:\,f(H)=\{0\}\}$. In this commutative case, if $f\in J$, then so is $S(f)$. If $J$ is proper, then $\varepsilon(f)=0$.

In Remark 2.10 of this paper (p.667), Wang says that for noncommutative $F(G)$ we still have $S(J)\subset J$ and $\varepsilon(J)=\{0\}$ for $J$ a Woronwicz $\mathrm{C}^*$-ideal ($J$ must be proper for this second condition to hold).

I can show using the antipodal property that if $S(J)\subset J$ then $\varepsilon(J)=\{0\}$... but $S(J)\subset J$ is eluding me.

Question:How does $S(J)\subset J$ follow?

**Some Efforts:**

We have that $$F(G)\otimes \ker \pi+\ker\pi\otimes F(G)\subset \ker(\pi\otimes \pi),$$ and I believe that due to finite dimensionality

$$\ker(\pi\otimes\pi)=F(G)\otimes \ker \pi+\ker\pi\otimes F(G).$$

This means that if $j\in J$, and $$\Delta(j)=\sum_i j_{(1),i}\otimes j_{(2),i},$$ that for each $i$, either $j_{(1),i}$ or $j_{(2),i}\in J$.

If we write $j\in J$ in terms of the matrix elements of irreducible unitary representations: $$j=\sum_\underset{\alpha\in\text{Irr}(G)}{i,j=1}^{d_\alpha}a_{ij}^{\alpha}\rho_{ij}^\alpha,$$ that $$j^*=\sum_\underset{\alpha\in\text{Irr}(G)}{i,j=1}^{d_\alpha}\overline{a_{ij}^{\alpha}}S(\rho_{ji}^\alpha)\in J,$$ as $J$ is a $\mathrm{C}^*$-algebra and $S(\rho_{ij}^\alpha)=(\rho_{ji}^\alpha)^*$.

Any help would be appreciated... I just fear there is something simple here that I am missing.