In a textbook I found the following exercise:
Let $V = V_1 \oplus V_2 \oplus \cdots \oplus V_n$ and $W = W_1 \oplus W_2 \oplus \ldots \oplus W_n$ and $A_i \in \mbox{hom}(V_i, W_i)$ for $i = 1,\ldots n$. Show that there exists a unique transformation $A : V \to W$ such that $$ A(v_1 + v_2 + \ldots + v_n) = A_1v_1 + A_2v_2 + \ldots + A_nv_n. $$ for $v_i \in V$. For me the solution is quite obvious, suppose we have two such transformations $A,B \in \mbox{hom}(V,W)$. For some $v \in V$ we have a unique decomposition $v = v_1 + v_2 + \ldots + v_n$ as a property of the direct sum, so that $$ B(v) = B(v_1 + v_2 + \ldots + v_n) = A_1 v_1 + A_2 v_2 + \ldots + A_n v_n = A(v_1 + v_2 + \ldots + v_n) = Av $$ implying $Bv = Av$ for every $v \in V$, showing that $B = A$. Now I asked myself if every $A \in \mbox{hom}(V,W)$ could be decomposed in such $A_i \in \mbox{hom}(V_i, W_i)$, i.e. for every $A \in \mbox{hom}(V,W)$ there exists $A_i : V_i \to W_i$ such that $$ A(v_1 + v_2 + \ldots + v_n) = A_1v_1 + A_2v_2 + \ldots + A_nv_n $$ for $v_i \in V_i$. I guess, as every $v_i \in V_i$ corresponds to a unique vector $v_i \in V$, we can define $$ A_iv_i := P_i Av_i $$ where $P_i : V \to V_i$ denotes the unique projection onto the $i$-the space $V_i$. This gives a mapping $A_i : V_i \to W_i$. Then $$ A(v_1 + v_2 + \ldots + v_n) = Av_1 + Av_2 + \ldots + Av_n $$ but $Av_i \in W$ and here I am stuck. So my question, is such an decomposition possible and am I on the right track?