Let $\Delta$ be a commutative monoid and $S$ a magma. Assume $\varphi:S\rightarrow\Delta$ is a homomorphism. Define $S_\lambda:=\varphi^{-1}(\{\lambda\})$ for all $\lambda\in\Delta$. Clearly $S_{\lambda}S_\mu\subset S_{\lambda+\mu}$ for all $\lambda,\mu$. Moreover the $S_\lambda$'s are pairwise disjoint.
Let $(A_\lambda)_{\lambda\in\Delta}$ be a grading of a commutative ring $A$. Let $A^{(S)}$ denote the subset of $A^S$ consisting of elements of finite support. Write $\phi$ for the canonical magma homomorphism of $S$ into $A^{(S)}$.
For each $\gamma\in\Delta$, let $V_\gamma:=\{(\alpha,s)\in\bigcup_{\lambda\in\Delta}A_\lambda\times S\ |\ \deg(\alpha)+\varphi(s)=\gamma\}$ and set $E_\gamma:=\text{span}_{\mathbf{Z}}((\alpha\phi_s)_{(\alpha,s)\in V_\gamma})$.
I want to show that $\sum_{\gamma\in\Delta}E_\gamma$ is direct.
Let $x_\gamma\in E_\gamma$ for each $\gamma\in\Delta$ such that $\sum_{\gamma\in\Delta}x_\gamma=0$. There exists $\xi^\gamma\in\mathbf{Z}^{(V_\gamma)}$ such that $x_\gamma=\sum_{(\alpha,s)\in V_\gamma}\xi^\gamma_{\alpha s}(\alpha\phi_s)$. This implies that $$\sum_{\gamma\in\Delta}\sum_{(\alpha,s)\in V_\gamma}\xi^\gamma_{\alpha s}(\alpha\phi_s)=0.$$ How can I show that this implies that $\sum_{(\alpha,s)\in V_\gamma}\xi^\gamma_{\alpha s}(\alpha\phi_s)=0$ for all $\gamma\in\Delta$?
EDIT: Maybe the difficulty arises from the way I have defined the $E_\gamma$? I want $E_\gamma$ to be the additive group of $A^{(S)}$ generated by the elements $\alpha\phi_s$ such that $\alpha\in A_\lambda$, $s\in S_\lambda$ and $\lambda+\mu=\gamma$. Is there a better way to define $E_\gamma$?