Let $A$ be a ring and $M,N$ be two graded $A$-modules of type $\Delta$. For $\lambda\in\Delta$, let $X_\lambda:=\{f\in\text{Hom}_A(M,N)\ |\ \deg(f)=\lambda\}$. Write $\text{Homgr}_A(M,N):=\sum_{\lambda\in\Delta}X_\lambda$. Clearly this sum is direct.
Suppose $M$ is finitely generated. I want to show that $\text{Homgr}_A(M,N)=\text{Hom}_A(M,N)$.
Attempt:
There exists a mapping $x:I\rightarrow\bigcup_{\lambda\in\Delta}M_\lambda$ for some finite set $I$ such that $(x_i)_{i\in I}$ generated the $A$-module $M$. For each $i\in I$, let $\deg(x_i)=:d_i$. At this point the author claims the following
Let $u\in\text{Hom}_A(M,N)$ and for all $\lambda\in\Delta$ let $z_{i,\lambda}$ denote the homogeneous component of $u(x_i)$ of degree $\lambda+d_i$. We show that there exists an $A$-linear mapping $u_\lambda:M\rightarrow N$ such $u_\lambda(x_i)=z_{i,\lambda}$ for all $i\in I$. It suffices to prove that if $\sum_{i\in I}a_ix_i=0$, for $a_i\in A$, then $\sum_{i\in I}a_iz_{i\lambda}=0$ for all $\lambda\in\Delta$.
How does the fact that $\sum_ia_ix_i=0$ implies $\sum_ia_iz_{i\lambda}=0$ imply that $u_\lambda$ exists?