We will first define some basic cochain complexes which model the $n$-disk and $n$-sphere. $D(n)$ is the cochain complex generated by one element $b \in D(n)^n$ and its differential $c = d(b) \in D(n)^{n+1}$. On the other hand we define $S(n)$ to be the cochain complex generated by one element $a \in S(n)^n$ with trivial differential (i.e. $d a = 0$). In other words:
$$ D(n) = ... \to 0 \to \k \to \k \to 0 \to ... $$
$$ S(n) = ... \to 0 \to \k \to 0 \to 0 \to ... $$

Note that $D(n)$ is acyclic for all $n$, or put in different words: $j_n : 0 \to D(n)$ induces an isomorphism in cohomology. The sphere $S(n)$ has exactly one non-trivial cohomology group $H^n(S(n)) = \k \cdot [a]$. There is an injective function $i_n : S(n+1) \to D(n)$, sending $a$ to $c$. The maps $j_n$ and $i_n$ play the following important role in the model structure of cochain complexes:

\todo{Introduceer de model structuur}
\begin{claim}
	The set $I = \{i_n : S(n+1) \to D(n) \I n \in \N\}$ generates all cofibrations and the set $J = \{j_n : 0 \to D(n) \I n \in \N\}$ generates all trivial cofibrations.
\end{claim}

As we do not directly need this claim, we omit the proof. However, in the next section we will prove a similar result for cdga's in detail.

$S(n)$ plays a another special role: maps from $S(n)$ to some cochain complex $X$ correspond directly to elements in the kernel of $\restr{d}{X^n}$. Any such map is null-homotopic precisely when the corresponding elements in the kernel is a coboundary. So there is a natural isomorphism: $\Hom(S(n), X) / {\simeq} \iso H^n(X)$.

By using the free cdga functor we can turn these cochain complexes into cdga's $\Lambda(D(n))$ and $\Lambda(S(n))$. So $\Lambda(D(n))$ consists of linear combinations of $b^k$ and $c b^k$ when $n$ is even, and it consists of linear combinations of $c^k b$ and $c^k$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
$$ d(b^k) = k \cdot c b^{k-1} $$
$$ d(c b^k) = 0 $$

$$ d(c^k b) = c^{k+1} $$
$$ d(c^k) = 0 $$

Those cocycles are in fact coboundaries (using that $\k$ is a field of characteristic $0$):
$$ c b^k = \frac{1}{k} d(b^{k+1}) $$
$$ c^k = d(b c^{k-1}) $$

There are no additional cocycles in $\Lambda(D(n))$ besides the constants and $c$. So we conclude that $\Lambda(D(n))$ is acyclic as an algebra. In other words $\Lambda(j_n): \k \to \Lambda D(n)$ is a quasi isomorphism.

The situation for $\Lambda S(n)$ is easier as it has only one generator (as algebra). For even $n$ this means it is given by polynomials in $a$. For odd $n$ it is an exterior algebra, meaning $a^2 = 0$. Again the sets $\Lambda(I) = \{ \Lambda(i_n) : \Lambda S(n+1) \to \Lambda D(n) \I n \in \N\}$ and $\Lambda(J) = \{ \Lambda(j_n) : \k \to \Lambda D(n) \I n \in \N\}$ play an important role.

\begin{theorem}
	The sets $\Lambda(I)$ and $\Lambda(J)$ generate a model structure on $\CDGA_\k$ where:
	\begin{itemize}
		\item weak equivalences are quasi isomorphisms,
		\item fibrations are (degree wise) surjective maps and
		\item cofibrations are maps with the left lifting property against trivial fibrations.
	\end{itemize}
\end{theorem}

We will prove this theorem in the next section. Note that the functors $\Lambda$ and $U$ thus form a Quillen pair with this model structure.

\subsection{Why we need $\Char{\k} = 0$ for algebras}
The above Quillen pair $(\Lambda, U)$ fails to be a Quillen pair if $\Char{\k} = p \neq 0$. We will show this by proving that the maps $\Lambda(j_n)$ are not weak equivalences for even $n$. Consider $b^p \in D(n)$, then by the Leibniz rule:
$$ d(b^p) = p \cdot c b^{p-1} = 0. $$
So $b^p$ is a cocycle. Now assume $b^p = d x$ for some $x$ of degree $p n - 1$, then $x$ contains a factor $c$ for degree reasons. By the calculations above we see that any element containing $c$ has a trivial differential or has a factor $c$ in its differential, contradicting $b^p =  d x$. So this cocycle is not a coboundary and $\Lambda D(n)$ is not acyclic.