Given two graded vector spaces $A = (a_1, a_2, \ldots, a_n)$ and $B = (b_1, b_2, \ldots, b_n)$, and a lie algebra over each of them, $L(A)$ and $L(B)$, we can define a coproduct of $L(A)$ and $L(B)$, $L(A)\sqcup L(B)$. By the definition of a coproduct this means it must follow the rules of coproducts, so there must be inclusion morphisms $i_1: L(A)\to $L(A)\sqcup L(B)$ and $i_2: L(B)\to $L(A)\sqcup L(B)$ such that for any $f_1: L(A)\to \mathcal M$ and $g_2: L(B) \to \mathcal M$, there must exist a unique $f: L(A)\sqcup L(B) \to \mathcal M$ such that $f_1 = f\circ i_1$ and $g_2 = g\circ i_2$. And there must be a bracket on $L(A)\sqcup L(B)$ with the usual behavior.
Is it ever possible to construct a bracket for $L(A)\sqcup L(B)$ such that $[i_1(a_1), i_2(b_1)]\ne 0$? If I blindly take the usual definition of a direct sum for $L(A)$ and $L(B)$, such a term must be zero.