Let $V$ be a vector space with a basis $v_1, v_2, \ldots, v_n$. Let $T(V)$ be the tensor algebra of $V$. Let $S(Lie(V))$ be the symmetric algebra of the free Lie algebra of $V$. I think that $T(V)$ is isomorphic to $S(Lie(V))$ as a graded vector space. For example, let $n=2$. We have the degree $2$ component of $T(V)$ is \begin{align} T(V)_2 = Span\{v_1 \otimes v_1, v_1 \otimes v_2, v_2 \otimes v_1, v_2 \otimes v_2\}. \end{align} The degree 2 component of $S(Lie(V))$ is \begin{align} S(Lie(V))_2 = Span\{v_1^2, v_2^2, v_1 v_2, [v_1, v_2]\}. \end{align} A bijection from $T(V)_2$ to $S(Lie(V))_2$ is \begin{align} & v_1 \otimes v_1 \mapsto v_1^2, \\ & v_1 \otimes v_2 \mapsto v_1 v_2, \\ & v_2 \otimes v_1 \mapsto [v_1, v_2], \\ & v_2 \otimes v_2 \mapsto v_2^2. \end{align}
The degree $3$ component of $T(V)$ is \begin{align} & T(V)_3 = Span\{v_1 \otimes v_1 \otimes v_1, v_1 \otimes v_1 \otimes v_2, v_1 \otimes v_2 \otimes v_1, v_2 \otimes v_1 \otimes v_1, \\ & \quad \quad \quad \quad \quad v_1 \otimes v_2 \otimes v_2, v_2 \otimes v_1 \otimes v_2, v_2 \otimes v_2 \otimes v_1, v_2 \otimes v_2 \otimes v_2 \}. \end{align} The degree 2 component of $S(Lie(V))$ is \begin{align} S(Lie(V))_2 = Span\{v_1^3, v_2^3, v_1^2 v_2, v_1 v_2^2, v_1 [v_1, v_2], v_2[v_1,v_2], [[v_1,v_2],v_2], [[v_1,v_2],v_1]\}. \end{align} A bijection from $T(V)_3$ to $S(Lie(V))_3$ is \begin{align} & v_1 \otimes v_1 \otimes v_1 \mapsto v_1^3, \\ & v_2 \otimes v_2 \otimes v_2 \mapsto v_2^3, \\ & v_1 \otimes v_2 \otimes v_2 \mapsto v_1 v_2^2, \\ & v_1 \otimes v_1 \otimes v_2 \mapsto v_1^2 v_2, \\ & v_1 \otimes v_2 \otimes v_1 \mapsto v_1 [v_1, v_2], \\ & v_2 \otimes v_1 \otimes v_2 \mapsto [v_1, v_2]v_2, \\ & v_2 \otimes v_2 \otimes v_1 \mapsto [[v_1,v_2],v_2], \\ & v_2 \otimes v_1 \otimes v_1 \mapsto [[v_1, v_2], v_1]. \end{align} How to write down the general formula $T(V)_n \to S(Lie(V))_n$?