Proving the Eichler-Shimura Isomorphism defines a global section Let  $ f \in S_k(\Gamma)$ be a weight k modular cusp form of level $\Gamma$, with modular curve $Y_{\Gamma}$. Let $V^{k-2}$ be the homogenous polynomials in X and Y of degree k-2 with complex coefficients. Let $\Omega^1$ be the sheaf of differential 1 forms. We define the map:
$\phi : S_k(\Gamma) \to H^0(Y_{\Gamma}, \Omega^1 \otimes V^{k-2})$
$\phi(f) : = f(z)(X-zY)^{k-2}dz$
Question: How do I show that this defines a global section of $Y_{\Gamma}$? There must be some argumentation about the coordinate charts of the modular curve.

Note: This helps give an answer to my previous question (Eichler-Shimura Isomorphism) Proving c(f) is not a boundary) via the following observations:
By Dolbeaut's theorem, $H^0(Y_{\Gamma}, \Omega^1 \otimes V^{k-2} )\cong H^1(Y_\Gamma, V^{k-2})$. Then, since $Y_{\Gamma}$ is a $K(\Gamma, 1)$ surface, we know $H^1(Y_{\Gamma}, V^{k-2}) \cong H^1(\Gamma, V^{k-2})$, which is group cohomology, and the subject of my previous post. By showing that this $f$ defines a global section, it is trivial to show it is not a boundary in cohomology, via the isomorphisms.
 A: What exactly is a global section of $\Omega^1 \otimes V^{k-2}$ on $Y_\Gamma$?  First consider $\Omega^1 \otimes V^{k-2}$ as a sheaf on $\mathbb{H}$.  This sheaf is naturally endowed with a $\Gamma$ action through $\Gamma$'s action on $V^{k-2}$.
Cover $\mathbb{H}$ with open sets $\{U_i\}$ such that $\gamma(U_i)\cap U_I=\varnothing$.  For each $U_i$ we choose a local section $f_i\in\Gamma(U_i, \Omega^1 \otimes V^{k-2})$ that is "compatible" under the action of $\Gamma$ in the following sense: For any open set $V\subset \gamma(U_i) \cap U_j$, the pullback of $f_j|_V$ along $\gamma$ gives $\gamma(f_i)|_{\gamma^{-1}(V)}$ (note that by $\gamma(f_i)$ I mean the action of $\gamma$ on $f_i$).
In our particular situation, set $\gamma =\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and compute
\begin{eqnarray}
\gamma^* (f(z)(X-zY)^{k-2}dz) &=& f(\gamma(z))(X-\gamma(z)Y)^{k-2}d\gamma(z) \\
&=&(cz+d)^kf(z)(X-\frac{az+b}{cz+d}Y)^{k-2}(cz+d)^{-2}dz \\
&=&f(z)\gamma((X-zY)^{k-2}) dz. 
\end{eqnarray}
When $k=2$, the $(cz+d)^{-2}$ that comes out of pulling back $dz$ is exactly what is needed to cancel out the factor of automorphy.  When $k>2$, we need to add a "twisting" factor that will compensate.   
