Let $V$ be a graded vector space, thought of as a collection $\{ V^n \}_{n \ge 0}$ of vector spaces. Let $V_{odd} = \bigoplus_{n \text{ odd}} V^n$ and $V_{even} = \bigoplus_{n \text{ even}} V^n$. I am trying to show that there is a functor $$ \Lambda \colon \mathsf{GVec} \longrightarrow \mathsf{CGAlg} $$ between graded vector spaces, and commutative graded algebras which is left adjoint to the forgetful functor - i.e. $\Lambda V$ should be the free commutative graded algebra on $V$.
One way to define this is $\Lambda V = E( V_{odd} ) \otimes S(V_{even} )$ where $E$ denotes the exterior algebra and $S$ the symmetric algebra. However I am having trouble showing that it is commutative.
We can think of the exterior and symmetric algebras as polynomial algebras on basis elements of each of $V_{odd}$, $V_{even}$ respectively, subject to some relations. Let $\underline{x} \otimes \underline{y}, \underline{z} \otimes \underline{w}$ be elements in $\Lambda V \otimes \Lambda V$, where the underline means a monic monomial. Multiplying these gives $$ (\underline{x} \otimes \underline{y} ) \cdot (\underline{z} \otimes \underline{w}) = (-1)^{|\underline{y}| | \underline{z}|} \underline{x} \underline{z} \otimes \underline{y} \underline{w} \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ by definition of the tensor product of graded algebras. However if we apply the braid to the element $\underline{x} \otimes \underline{y}\otimes \underline{z} \otimes \underline{w}$ this gives $$ (-1)^{| \underline{x} \otimes \underline{y}||\underline{z} \otimes \underline{w}| } \underline{z} \otimes \underline{w} \otimes \underline{x} \otimes \underline{y} = (-1)^{(| \underline{x}| + |\underline{y}|)(|\underline{z}\ + |\underline{w}| )} \underline{z} \otimes \underline{w} \otimes \underline{x} \otimes \underline{y} $$ then multiply we get $$ (-1)^{(| \underline{x}| + |\underline{y}|)(|\underline{z}| + |\underline{w}| )} (-1)^{|\underline{w}| |\underline{x}|}\underline{z} \underline{x}\otimes \underline{w} \underline{y} $$ We can swap $\underline w$ and $\underline y$ since they live in the symmetric algebra, and can swap $\underline z$ and $\underline x$ after multiplying by $(-1)^{| \underline x| |\underline z | }$ so the above expression simplifies to $$ (-1)^{|\underline y| |\underline z| + |\underline y| |\underline w |} \underline{x} \underline{z}\otimes \underline{y} \underline{w} \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$
(1) and (2) should be the same. What have I done wrong?
EDIT: Just to be clear I am trying to show that $\mu \sigma = \mu$ where $\sigma$ in $\mathsf{GVec}$ is the braiding $a \otimes b \mapsto (-1)^{|a||b|}b \otimes a$ and $\mu$ is the multiplication.