I am reading the following paper of Zagier, and I am confused as to his identification of parabolic 1-cocyles and representations.
We will let $\Gamma=PSL_2(\mathbb{Z})$, and we shall denote $S=\begin{bmatrix}0 & -1\\ 1 &0\end{bmatrix}$, $U=\begin{bmatrix}1 & -1\\ 1&0\end{bmatrix}$, then $T=US=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$. We recall that $S^2=U^3=I$. Now if we let $V$ be a right representation of $\Gamma$ whose action is denoted $v\mapsto v\vert\gamma$, then we define the parabolic 1-cocycles as $$ Z_0^1(\Gamma,V)=\{f:\Gamma\rightarrow V:f(T)=0,f(\gamma_1\gamma_2)=f(\gamma_1)\vert \gamma_2+f(\gamma_2)\} $$ Now Zaiger says that by the map $f\mapsto f(S)$, we can identify this with the space $$ W=\{v\in V: v\vert(1+S)=v\vert(1+U+U^2)=0\} $$ I am trying to understand this identification. I see that given such a function $f$ if we send it to $f(S)\in V$. Then using the cocycle relation we will have that $$ v\vert(1+S)=f(S)\vert(1+S)=f(S(1+S))-f(1+S) $$ $$ =f(S+S^2)-f(1+S)=f(S+1)-f(1+S)=0 $$ where we have used the fact that $S^2=1$. However, I am having a harder time deducing that $v\vert(1+U+U^2)=0$. I have tried several things along the lines of multiplying out the matrices and seeing if things reduce. I know I will likely have to use that $f(T)=0$ at some point, but the only place where I have gotten an $f(T)$ is after expanding with the cocycle condition, I get a $f(SU^2)=f(S^{-1}U^{-1})=f((US)^{-1})=f(T^{-1})=0$, but the rest of the terms aren't canceling out. I'm sure it is just some trick I am not seeing, so any help would be appreciated.
Furthermore, I am curious as to what the inverse of this identification would be that is given a $v\in V$ what is the 1-cocycle that I can associate with it? Any and all help is appreciated.