Coalgebra differential on reduced symmetric algebra Definitions/setup
(I am using "Formality and star products" by Cattaneo)
Let $g=\bigoplus_{i\in \mathbb{Z}} g^i$ be a graded vector space over some field. We can then look at the reduced symmetric space $\overline{S}(g[1])$ (where $g[1]=\bigoplus_{i\in \mathbb{Z}} g^{i+1})$ which is defined by dividing the tensor algebra $\overline{T}(g[1])=\bigoplus_{n=1}^\infty g[1]^{\otimes n}$ by the ideal generated by elements of the form $x\otimes y-T(x\otimes y)$ where $T$ is the twisting map. This graded vector space is made into an algebra using tensor products as multiplication.
We can make this graded symmetric tensor algebra into a graded coalgebra by defining the comultiplication $\Delta(v)=1\otimes v +v\otimes 1$ on $v \in g[1]$ and extend this using tensor products.
My question
Suppose now that we have a degree 1 coalgebra differential on $\overline{S}(g[1])$ say $Q$. In the document "Formality and starproducts" it is stated after definiton 3.6 that it is enough to know $Q^1$ which is the composition of $Q$ and the projection onto the first component. I dont understand this, because since $Q$ is a degree 1 differential, it means that it sends pure tensors of degree $n$ to tensors of degree $n+1$, this means that for any element in $\overline{S}(g[1])$ the first component of the image under $Q$ will be $0$. 
Clearly, I must be missing something here. I hope someone can help me.
 A: There are two gradings on $\bar{S}(g[1])$. The first is the usual degree grading, given by $\deg(x_1 \otimes \dots \otimes x_n) = \deg(x_1) + \dots + \deg(x_n)$ where $x_i \in g[1]$. The second, also commonly called weight, is given by $w(x_1 \otimes \dots \otimes x_n) = n$ (the number of "letters" in the "word", if you want).
The $Q^1$ they mention is the projection on the first weight component $\bar{S}^1(g[1]) = g[1]$. More generally define $\bar{S}^n(g[1])$ to be the image of $(g[1])^{\otimes n}$ in the quotient, then
$$\bar{S}(g[1]) = \bigoplus_{n=1}^\infty \bar{S}^n(g[1]),$$
and $Q^1 : \bar{S}(V) \to V$ is the composite of $Q$ with the projection $\operatorname{pr}_1 : \bar{S}(V) \to \bar{S}^1(V) = V$
It's true that $Q$ is a degree $1$ differential, so $\deg(Q(x)) = \deg(x) + 1$. But even if $x$ is a "word with $n$ letters" (i.e. has weight $n$) then $Q(x)$ will typically be a sum of monomials of varying lengths, not necessarily just $n$, or $n+1$ letters.

Now why is $Q$ uniquely determined by $Q^1$? $\bar{S}(g[1])$ is what is called a cofree cocommutative coalgebra (more precisely, it's a cofree conilpotent cocommutative coalgebra). This is a notion formally dual to the notion of a free commutative algebra, that is, a polynomial algebra $S(V)$.
For ease of notation I'll consider some graded vector space $V$ (that you can specialize as $g[1]$ later). It is well known that a given derivation $d : S(V) \to S(V)$ (i.e. a map satisfying $d(ab) = d(a)b \pm a d(b)$) is uniquely determined by its restriction $f$ on $V = S^1(V) \subset S(V)$, because if you have a monomial $v_1 \dots v_n \in S(V)$, you can use the derivation relation to compute
$$d(v_1 \cdot \dots \cdot v_n) = \sum_{i=1}^n \pm v_1 \dots f(v_i) \dots v_n.$$
What's more, given any linear map $f : V \to S(V)$, you get a uniquely defined derivation $d_f : S(V) \to $ with the above formula.
Well, since the notion of cofree coalgebra is formally dual to the notion of free algebra, it's not surprising that a coderivation $Q : \bar{S}(V) \to \bar{S}(V)$ is uniquely determined by its "corestriction" to $\bar{S}^1(V)$, i.e. the composite $\operatorname{pr}_1 \circ Q : \bar{S}(V) \to \bar{S}(V) \to \bar{S}^1(V) = V$. If you are not used to working with coalgebras, it might be harder to see, but I'll try to explain.
Recall that a coderivation $Q$ is characterized by the relation $\Delta Q = (Q \otimes 1 + 1 \otimes Q) \Delta$. Suppose that you have some coderivation $Q : \bar{S}(V) \to \bar{S}(V)$ and that you know how to compute the corestriction $Q^1$. Write $Q = Q^1 + Q^2 + \dots$, where $Q^n$ is the corestriction to $\bar{S}^n(V)$. How do you compute $Q^2$? Take some $x \in \bar{S}(V)$ and write $\Delta(x) = \sum_{(x)} x_{(1)} \otimes x_{(2)}$ (I'm using Sweedler's notation). Then the coderivation relation tells you:
$$\Delta(Q(x)) = \sum_{(x)} \bigl( Q(x_{(1)}) \otimes x_{(2)} \pm x_{(1)} \otimes Q(x_{(1)}) \bigr) \in \bar{S}(V) \otimes \bar{S}(V).$$
Now corestrict (project) both parts of the tensor product onto $\bar{S}^1(V)$. What does it tell you?


*

*On the one hand, because of how the coproduct of $\bar{S}(V)$ is defined, the projection of the LHS is precisely $Q^2(x)$. Indeed, if $n \ge 3$, then at least one of the factors of $\Delta(Q^n(x)) \in \bar{S}(V) \otimes \bar{S}(V)$ will have at least weight $2$, and so $(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(Q^n(x)) = 0$. And $\Delta(Q^1(x)) = 0$ anyway, so all you're left is $Q^2$, and by definition on a polynomial of weight $2$ you just get
$$(\operatorname{pr}_1 \otimes \operatorname{pr}_1) \Delta(v \cdot w) = v \otimes w.$$

*On the other hand, the projection of the RHS can be expressed in terms of $Q^1$, and so you get:
$$Q^2(x) = \sum_{(x)} \bigl( Q^1(x_{(1)}) \otimes x_{(2)} \pm x_{(1)} \otimes Q^1(x_{(1)}) \bigr).$$


(More precisely, $Q^2$ will be the projection of this in the quotient by the transpose map $T$.) A similar reasoning gives:
$$Q^n(x) = \sum_{(x)} \sum_{i=1}^n \pm x_{(1)} \otimes \dots \otimes Q^1(x_{(i)}) \otimes \dots \otimes x_{(n)},$$
and so you can recover the whole coderivation just from $Q^1$. And as before, any linear map $f : \bar{S}(V) \to V$ defines a coderivation with the above formula (where $Q^1 = f$).
