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 remember the topological $n$-simplex defined as convex span.
$$(\Apl)_n =\frac{\Lambda(x_0, \ldots, x_n, d x_0, \ldots, d x_n)}{(\sum_{i=0}^n x_i -1, \sum_{i=0}^n d x_i)}, $$
where $\deg{x_i}=0$. 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 $\sum_{i=0}^n d x_i =0$.
Note that the inclusion $\Lambda(x_1, \ldots, x_n, d x_1, \ldots, d x_n)\to\Apl_n$ 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 $d x_i$). This fact will be used throughout. Also note that we have already seen the dual unit interval $\Lambda(t, dt)$ which is isomorphic to $\Apl_1$.
One can check that $\Apl\in\simplicial{\CDGA_\k}$. We will denote the subspace of homogeneous elements of degree $k$ as $\Apl^k$, this is 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 $s: \Apl_n \to\Apl_{n+1}$. In the more geometric context of topological $n$-simplices we would achieve this by dividing by $1-x_0$. However, since this algebra consists of polynomials only, this cannot be done. Instead, we will multiply everything by $(1-x_0)^2$, so that we can divide by $1-x_0$. Define for $i =1, \ldots, n$:
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 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 $\Apl^k \to\ast$ is a weak equivalence.
This is clear for $\Apl_0=\k\cdot1$. For $\Apl_1$ we see that $\Apl_1=\Lambda(x_1, d x_1)\iso\Lambda D(0)$, which we proved to be acyclic in the previous section.
For general $n$ we can identify $\Apl_n \iso\bigtensor_{i=1}^n \Lambda(x_i, d x_i)$, because $\Lambda$ is left adjoint and hence preserves coproducts. By the Künneth theorem \TheoremRef{kunneth} we conclude $H(\Apl_n)\iso\bigtensor_{i=1}^n H \Lambda(x_i, d x_i)\iso\bigtensor_{i=1}^n H \Lambda D(0)\iso\k\cdot[1]$.