This is taken from Timmermann's Invitation to Quantum Groups and Duality.
Hi folks I am struggling a little with a small calculation in the above text.
I will just get right into it.
Lemma 3.2.5 Let $(A,\Delta)$ be a Hopf $*$-algebra with a normalised integral $h$, let $\chi_V$ and $\chi_W$ be corepresentations on finite-dimensional vector spaces $V$ and $W$, respectively, and let $Q\in \hom(V,W)$. Denote by $X$ and $Y$ the corepresentation operators corresponding to $\chi_V$ and $\chi_W$, respectively, and define $R,T\in\hom(V,W)$ by $$R:=(I\otimes h)(Y^{-1}(Q\otimes 1)X)\,\,\,\textit{ and }\,\,\, T:=(I\otimes h)(Y(Q\otimes 1)X^{-1}).$$ Then $R,T\in\hom(\chi_V,\chi_W)$. If $Q\in\hom(\chi_V,\chi_W)$, then $R=T=Q$.
Proof Since $Y$ and $X$ are corepresentations, the composition
$$\begin{align} Y^{-1}(R\otimes 1)X&=Y^{-1}(I\otimes h)(Y^{-1}(Q\otimes 1)X)X \\&=(I\otimes h\otimes I)\left(Y_{[13]}^{-1}Y_{[12]}^{-1}(Q\otimes 1\otimes 1)X_{[12]}X_{[13]}\right) \end{align}$$
Question: Where does the second equality come from?
The Rest of the Proof
This second expression can be rewritten in the form $$(I\otimes h\otimes I)((I\otimes \Delta)\left(Y^{-1}\right)(Q\otimes 1\otimes 1)(I\otimes\Delta)(X)).$$
By right-invariance of $h$, this composition is equal to
$$(I\otimes h)\left(Y^{-1}(Q\otimes 1)X\right)\otimes 1=R\otimes 1.$$
Thus $(R\otimes 1)X=Y(R\otimes 1)$. A similar calculation shows that $(T\otimes 1)X=Y(T\otimes 1)$ (using left-invariance of $h$). A previous result shows that $R$ and $T$ are then elements of $\hom(\chi_V,\chi_W)$.
If $Q\in\hom(\chi_V,\chi_W)$, then $Y^{-1}(Q\otimes 1)X=Q\otimes 1=Y(Q\otimes 1)X^{-1}$ by that previous result and so $R=Q=T$.
Further Details
Normalised Integral: A linear functional $h:A\rightarrow \mathbb{C}$ is a normalised integral if it is both left-and right-invariant:
$$(I\otimes h)\Delta(a)=h(a)1_{\mathcal{M}(A)}\,\,\,\text{ and }\,\,\,(h\otimes I)\Delta(a)=h(a)1_{\mathcal{M}(A)},$$
(where $\mathcal{M}(A)$ is the multiplier algebra of the quantum group $A$), as well as normalised
$$h(1_A)=1.$$
Corepresentation Operator: Where $\chi:V\rightarrow V\otimes A$ is a corepresentation of $A$ on a vector space $V$, the corepresentation operator associated with $\chi$ is an $X\in\hom(V)\otimes A$ such that $$X=\sum_{i,j}e_{ij}\otimes a_{ij},$$
where the $a_{ij}$ are elements of $A$ given by
$$\chi(e_j)=\sum_i e_i\otimes a_{ij},$$
so that $\chi(v)=X(v\otimes 1_A)$.
The inverse of $X$ is given by $X^{-1}=(I\otimes S)(X)$ where $S$ is the antipode.
Leg Notation: The leg notation is given by $X_{[12]}=X_0\otimes 1_A$ and $X_{[13]}=(I\otimes\tau)(X\otimes 1_A)$ where $\tau:A\otimes B\rightarrow B\otimes A$ is the flip map and we can show that $$(I\otimes\Delta )X=X_{[12]}X_{[13]}.$$
Intertwiners: An intertwiner is a map $\varphi:V\rightarrow W$ such that $$\chi_W\varphi=(\varphi\otimes 1_A)\chi_V.$$ The space of intertwiners from a corepresentation $\chi_V:V\rightarrow V\otimes A$ to a corepresentation $\chi_W:W\rightarrow W\otimes A$ is denoted by $\hom(\chi_V,\chi_W)$.
My Thoughts
...are few and far between. If anything I think it should be $$Y^{-1}(R\otimes 1)X=Y^{-1}((I\otimes h)(Y^{-1}(Q\otimes 1)X)\otimes 1)X$$
I had previously tried to unravel this lemma but Timmermann called $R$ by $S$ which I confused with the antipode eventually to experience "death by a million linear maps".
Further information available of course on request.
