The question concerns part of a theorem in the book Foundations of Quantum Group Theory, Shahn Majid (Cambridge University Press, 1995). More specifically, Theorem 2.3.4 (p.55-57) which I'll rewrite below in terms of what's impeding my progress through the book.
Let $(H, \mathcal{R})$ be a quasitriangular Hopf algebra and let $\chi$ be a counital 2-cocycle. Define: $$ U=\chi^{(1)}(S\chi^{(2)}), $$ where $S$ denotes the antipode. Show that $UU^{-1}=1$.
To start with, we define $U^{-1}=S(\chi^{-(1)})\chi^{-(2)}$. The proof as is given in the book goes as follows,
$$ UU^{-1}= \chi^{(1)}(S\chi^{(2)})(S\chi^{-(1)})\chi^{-(2)} $$ $$ \qquad\qquad\qquad\qquad\quad\,\,= \chi'^{-(1)}\chi^{(1)}(S(\chi'^{-(2)}_{(1)}\chi^{-(1)}\chi^{(2)}))\chi'^{-(2)}_{(2)}\chi^{-(2)}$$ $$ \,\,= \chi^{-(1)}_{(1)}(S\chi^{-(1)}_{(2)})\chi^{-(2)}$$ $$ =1. \quad\qquad\qquad\quad\,\,\,\,$$
The author then proceeds to say the following:
We inserted a copy of $\chi$, denoted $\chi'$, in a form that collapses via the antipode properties and $(\text{id}\otimes \epsilon)\chi'^{-1}=1$. We then applied the 2-cocycle condition in the form $(\Delta\otimes\text{id}\chi^{-1}=((\text{id}\otimes\Delta)\chi^{-1})\chi_{23}^{-1}\chi_{12}$ and $(\epsilon\otimes\text{id})\chi^{-1}=1.$
My problem is that no matter what I try, I can do nothing to arrive at the precise form given in the second row of the proof, and I definitely have no clue how the last line equates to $1$, even though most of the major steps have been written. I tried inserting several identity relations (several "ones") in super specific ways, but that brought me nowhere. I am also unsure which specific antipode properties are being used here.
Any help with filling in the gaps will be greatly appreciated.
The relevant properties of the antipode include $$ m\circ(S\otimes\mathrm{id})\circ \Delta = m\circ(\mathrm{id}\otimes S)\circ \Delta = \eta\circ \epsilon\ , $$ which says that $$ \sum S(x_{(1)})x_{(2)} = \sum x_{(1)}S(x_{(2)}) = \epsilon(x)1 $$ when applied to an element $x$, as well as the fact that $S$ is an algebra antihomomorphism, $S(xy) = S(y)S(x)$.
Now if $\chi$ is invertible and satisfies $(\mathrm{id}\otimes\epsilon)\chi = 1$, then its inverse also satisfies this relation. Expanding this using $\chi' = \chi'^{-(1)}\otimes \chi'^{-(2)}$ (I'll drop the sum from here on), we get $1 = \chi'^{-(1)}\epsilon(\chi'^{-(2)})$. Multiplying this on the left of $\chi^{(1)}(S\chi^{(2)})(S\chi^{-(1)})\chi^{-(2)}$, you get $$ \chi'^{-(1)}\epsilon(\chi'^{-(2)})\chi^{(1)}(S\chi^{(2)})(S\chi^{-(1)})\chi^{-(2)} $$ Since $\epsilon(\chi'^{-(2)})$ is a scalar, we can slide it to the right, just before $\chi^{-(2)}$. We can also rewrite it using the first antipode property quoted above, so we get $$ \begin{align} U U^{-1} & = \chi'^{-(1)}\chi^{(1)}(S\chi^{(2)})(S\chi^{-(1)})\epsilon(\chi'^{-(2)})\chi^{-(2)} \\ & = \chi'^{-(1)}\chi^{(1)}(S\chi^{(2)})(S\chi^{-(1)})S({\chi'^{-(2)}}_{(1)}){\chi'^{-(2)}}_{(2)}\chi^{-(2)} \end{align} $$ Next, using the antihomomorphism property, we can group the factors in the middle involving $S$ into one factor $S({\chi'^{-(2)}}_{(1)}\chi^{-(1)}\chi^{(2)})$. This reproduces the second line in the original proof.
As for the last line of the proof, it's also an application of the first antipode property. Applying that to the first two factors, we get ${\chi^{-(1)}}_{(1)} (S{\chi^{-(1)}}_{(2)})\chi^{-(2)} = \epsilon(\chi^{-(1)})\chi^{-(2)}$, which is equal to $1$ using the counital property $(\epsilon\otimes \mathrm{id})\chi^{-1} = 1$.