Timmerman's "An invitation to quantum groups and duality" corollary 3.2.8 Consider the following propositions of Timmerman's book "An invitation to quantum groups and duality":

I do not understand the equivalence
$$h(\mathfrak{C}(\delta_V)S(\mathfrak{C}(\delta_W)))=0 \iff\operatorname{Hom}(\delta_V, \delta_W)=0$$
in Timmerman's corollary 3.2.8.
Maybe we should require that $\delta_V$ and $\delta_W$ are irreducible corepresentations here?
In that case, the implication from right to left follows from prop 3.2.6, once we observe that by prop 3.2.2. $\delta_V \not\cong \delta_W$.
However, it is still not clear to me why the other implication should hold, i.e. why is
$$h(\mathfrak{C}(\delta_V)S(\mathfrak{C}(\delta_W)))=0 \implies\operatorname{Hom}(\delta_V, \delta_W)=0$$ true.
 A: Everything is correct. First, suppose that the corepresentations are indeed irreducible. As you say, one of the implications follows from Prop. 3.2.6. For the opposite one: Suppose that $\mathop{\rm Hom}(\delta_V,\delta_W)\neq 0$. By Schur's lemma (Prop. 3.2.2) this actually means that $\delta_V\simeq\delta_W$, so $\mathcal{C}(\delta_V)=\mathcal{C}(\delta_W)$. Without loss of generality, assume that the corepresentation $\delta_V$ is unitary (by Thm 3.2.1; actually, non-degenerate is enough) and denote by $u_{ij}$ its entries (i.e. $\delta_V\colon e_i\mapsto\sum_j e_j\otimes u_{ji}$). Then
$$\sum_j h(u_{ij}S(u_{ji}))=\sum_j h(u_{ij}u_{ij}^*)=h(1_A)=1\neq 0,$$
so at least one of the summands must be nonzero and hence $h(\mathcal{C}(\delta_V)S(\mathcal{C}(\delta_V)))\neq 0$.
Now if $\delta_V$ and $\delta_W$ are not irreducible, then by Thm 3.2.1 they decompose into irreducibles $V=\bigoplus V_i$, $W=\bigoplus W_i$. Then
$$\eqalign{\mathop{\rm Hom}(\delta_V,\delta_W)\neq 0\quad&\Leftrightarrow\quad\exists i,j\;\mathop{\rm Hom}(\delta_{V_i},\delta_{V_j})\neq 0\cr&\Leftrightarrow\quad\exists i,j\;h(\mathcal{C}(\delta_{V_i})S(\mathcal{C}(\delta_{W_j})))\neq 0\quad\Leftrightarrow\quad h(\mathcal{C}(\delta_{V})S(\mathcal{C}(\delta_{W})))\neq 0
}$$
