I'm going through Felix-Oprea-Tanré's book Algebraic models in geometry and there's a remark that I'm unable to prove.
In remark 2.9 they say that, if $(\Lambda V, d)$ is a Sullivan cdga and $d_\textrm{lin}: V \to V$ denotes its linear part, then $d_\textrm{lin}$ is a differential.
Denote by $\pi_\textrm{lin}: \Lambda V \to \Lambda V / (\mathbb{Q} \oplus \Lambda^{\geq 2} V) \cong V$ the projection of $\Lambda V$ into its linear part. Then $d_\textrm{lin} = \pi_{\textrm{lin}} \circ d$, so $d_\textrm{lin}^2 = 0$ if, and only if, $\pi_\textrm{lin} \circ d \circ \pi_\textrm{lin} \circ d = 0$. How can one show this (I tried direct computation and it didn't work...)? An indication rather than the full solution would be very appreciated!
Side question: They use the notation $d_0$ instead of $\pi_\textrm{lin}$, but intuitively speaking shouldn't $d_0$ denote the projection of $d$ into $\mathbb{Q}$, and use $d_1$ for the linear part of $\Lambda V$? In other words, is there a reason to denote the linear part of $d$ as $d_0$?
A way of seeing this is to note that derivations in a free commutative graded algebra $SV$ are in bijection with graded linear maps $V\to SV$,
$$\operatorname{Der}(SV) \cong \operatorname{Hom}(V,SV).$$
If $d:SV\to SV$ is a derivation, then the corresponding linear map is simply the restriction $f:V\hookrightarrow SV\xrightarrow{d}SV$. For the converse, given any graded linear map $f: V\to SV$, define a graded linear map $d_f:SV\to SV$ extending by $f$ by using Leibniz's rule. That is, define $$d_f(x_1\cdots x_n) = \sum_{i=1}^n (-1)^{|x_1|+\cdots + |x_{i-1}|} x_1\cdots f(x_i)\cdots x_n.$$
Under this point of view, a derivation $d$ on $SV$ decomposes into the sum of its homogeneous components with respect to monomial lenght. That is to say, if we define $S^iV$ as the linear span of length $i$ monomials on $V$ inside $SV$, then $SV = \bigoplus_{i\geq 0} S^iV$. A derivation $d:SV\to SV$ decomposes uniquely as a sum $d=d_0+d_1+\cdots$, where each $d_i$ is the unique extension as a derivation of a linear map $V\to S^{i+1}V$.
Now, the fact that $d^2=0$ can also be written in homogeneous components gives that $$0 = d^2 = (d_0+d_1+d_2 +\cdots)\circ (d_0+d_1+d_2 +\cdots) = d_0^2 + (d_1d_0+d_0d_1) + (d_0d_2 + d_ 1^2 +d_2d_0) + \cdots$$ Above, I grouped together the homogeneous components of $d^2$. Since $d^2=0$, each homogeneous component vanishes. In particular, $d_0^2$ vanishes and therefore $d_0$ is a differential.
Regarding your question about labeling the linear part $d_1$ rather than $d_0$, this is just a matter of convention. The reason in this case is that there is a useful spectral sequence attached to $(SV,d)$ whose zero page $E^0$ is $SV$, and whose differential on the zero page $d^0$ is precisely the linear part of $d$. Therefore, to make consistent the notation of the decomposition of the differential with this spectral sequence, it is less confusing if we shift the indices in the homogeneous decomposition.