We will now give a cdga model for the $n$-simplex $\Delta^n$. This then allows for simplicial methods. In the following definition one should be reminded of the topological $n$-simplex defined as convex span.
So it is the free cdga with $n+1$ generators and their differentials such that $\sum_{i=0}^n x_i =1$ and in order to be well behaved $\sum_{i=0}^n dx_i =0$.
Note that the inclusion $\Lambda(x_1, \ldots, x_n, dx_1, \ldots, dx_n)\to\Apl_n$ is an isomorphism of cdga's. So $\Apl_n$ is free and (algebra) maps from it are determined by their images on $x_i$ for $i =1, \ldots, n$ (also note that this determines the images for $dx_i$). This fact will be used throughout.
These cdga's will assemble into a simplicial cdga when we define the face and degeneracy maps as follows ($j =1, \ldots, n$):
$$ d_i(x_j)=\begin{cases}
x_{j-1}, &\text{ if } i < j \\
0, &\text{ if } i = j \\
x_j, &\text{ if } i > j
\end{cases}\qquad d_i : \Apl_n \to\Apl_{n-1}$$
$$ s_i(x_j)=\begin{cases}
x_{j+1}, &\text{ if } i < j \\
x_j + x_{j+1}, &\text{ if } i = j \\
x_j, &\text{ if } i > j
\end{cases}\qquad s_i : \Apl_n \to\Apl_{n+1}$$
One can check that $\Apl\in\simplicial{\CDGA_\k}$. We will denote the subspace of homogeneous elements of degree $k$ as $\Apl^k \in\simplicial{\Mod{\k}}$, this is indeed a simplicial $\k$-module as the maps $d_i$ and $s_i$ are graded maps of degree $0$.
We will prove this by defining an extra degeneracy $s: \Apl_n \to\Apl_{n+1}$. Define for $i =1, \ldots, n$:
\begin{align*}
s(1) &= (1-x_0)^2 \\
s(x_i) &= (1-x_0) \cdot x_{i+1}
\end{align*}
Extend on the differentials and multiplicatively on $\Apl_n$. As $s(1)\neq1$ this map is not an algebra map, however it well-defined as a map of cochain complexes. In particular when restricted to degree $k$ we get a linear map:
$$ s: \Apl^k_n \to\Apl^k_{n+1}. $$
Proving the necessary properties of an extra degeneracy is fairly easy. For $n \geq1$ we get (on generators):
So $d_{i+1} s = s d_i$. Similarly $s_{i+1} s = s s_i$. And finally for $n=0$ we have $d_1 s =0$.
So we have an extra degeneracy $s: \Apl^k \to\Apl^k$, and hence (see for example \cite{goerss}) we have that $\Apl^k$ is contractible. As a consequence $\Apl\to\ast$ is a weak equivalence.
Besides the simplicial structure of $\Apl$, there is also the structure of a cochain complex.
\Lemma{apl-acyclic}{
$\Apl_n$ is acyclic, i.e. $H(\Apl_n)=\k\cdot[1]$.
}
\Proof{
This is clear foor $\Apl_0=\k\cdot1$. For $\Apl_1$ we see that $\Apl_1=\Lambda(x_1, dx_1)\iso\Lambda D(0)$, which we proved to be acyclic in the previous section.
For general $n$ we can identify $\Apl_n \iso\bigtensor_{i=1}^n \Lambda(x_i, dx_i)$, because $\Lambda$ is left adjoint and hence preserves coproducts. By the Künneth theorem \TheoremRef{kunneth} we conclude $H(\Apl_n)\iso\bigtensor_{i=1}^n H \Lambda(x_i, dx_i)\iso\bigtensor_{i=1}^n H \Lambda D(0)\iso\k\cdot[1]$.