Let $V:=\oplus\_{j\in\mathbb{Z}}V\_j$ be a graded $\mathbb{F}$-vector space over
the field $\mathbb{F}$. The graded tensor product of graded vector spaces is given
by 


$V \otimes W:= \oplus\_{j\in \mathbb{Z}}\oplus\_{p+q=j}V\_p\otimes V\_q$
and for the graded vector space $\mathbb{F}[j]$, which is $\mathbb{F}$ in degree
$j$ and te zero vector space $\{0\}$ otherwise, the shift $V[j]$ is given by


$V[j]:=\mathbb{F}[j]\otimes V$


We then define a monoidal structure on the category of graded vector space,
(more or less) given by the rule on homogeneous elements 


$v\otimes w= (-1)^{deg(v)deg(w)}w\otimes v$


Then there is the decalage isomorphism


$
dec: V\_1[1]\otimes \cdots \otimes V\_n[1] \to (V\_1 \otimes \cdots \otimes V\_n)[n]
$


given by $dec(v\_1[1]\otimes \cdots \otimes v\_n[1])= 
(-1)^{\sum\_{j=1}^n(n-j)deg(v\_j)}(v\_1\otimes \cdots \otimes v\_n)[n]$.


Now in work on graded (stuff), it is frequently said, that this isomorphism
defines a natural isomorphism of the symmetric graded tensor-algebra of $V[1]$
and the antisymmetric graded tensor algebra, that is


$S(V[1])\simeq (\bigwedge V)[n]$


\**The question is: How does the decalage induces such an algebra isomorphism? Or
What is the natural isomorphism? \**


If $dec$ itself would be the isomorphism, then


$dec(v[1] \vee w[1])= (dec(v\_1)\wedge dec(w))[2]$ should hold, but this isn't true in general.



Natural Isomorphism of $S(V[1])$ and $(\bigwedge V)[n]$
You have a detailed proof (in a much more general context but easy to read) in Proposition I.4.3.2.1 of Illusie, Complexe Cotangent et Déformations I, Springer LNM 239. 


