In this section we will discuss the so called minimal models. These are cdga's with the property that a quasi isomorphism between them is an actual isomorphism.
\begin{definition}
An cdga $(A, d)$ is a \emph{Sullivan algebra} if
\begin{itemize}
\item$(A, d)$ is quasi-free (or semi-free), i.e. $A =\Lambda V$ is free as a cdga, and
Let $V$ generate $A$. Take $V(n)=\bigoplus_{k=0}^n V^k$ (note that $V^0= V^1=0$). Since $d$ is decomposable we see that for $v \in V^n$: $d(v)= x \cdot y$ for some $x, y \in A$. Assuming $dv$ to be non-zero we can compute the degrees:
Let $(A, d)$ be an $1$-connected cdga, then it has a minimal model.
\end{theorem}
\begin{proof}
We will construct a sequence of models $m_k: (M(k), d)\to(A, d)$ inductively.
\begin{itemize}
\item First define $V(0)= V(1)=0$ and $m_0= m_1=0$. Then set $V(2)= H^2(A)$ and define a map $m_2: V(2)\to A$ by picking representatives.
\item Suppose $m_k: (\Lambda V(k), d)\to(A, d)$ is constructed. Choose cocycles $a_\alpha\in A^{k+1}$ and $z_\beta\in(\Lambda V(k))^{k+2}$ such that $H^{k+1}(A)=\im(H^{k+1}(m_k))\oplus\bigoplus_\alpha\k\cdot[a_\alpha]$ (so $m_k$ together with $a_\alpha$ span $H^{k+1}(A)$) and $\ker(H^{k+2}(m_k))=\bigoplus_\beta\k\cdot[z_\beta]$. Note that $m_k z_\beta= db_\beta$ for some $b_\beta\in A$.
Define $V(k+1)=\bigoplus_\alpha\k\cdot v'_\alpha\oplus\bigoplus\k\cdot v''_\beta$ and set $dv'_\alpha=0$, $dv''_\beta= z_\beta$, $m_k(v'_\alpha)= a_\alpha$ and $m_k(v''_\beta)= b_\beta$.
\end{itemize}
This ends the construction. We will prove the following assertion for $k \geq2$:
$$ H^i(m_k)\text{ is }\begin{cases}
\text{an isomorphism}&\text{ if } i \leq k \\
\text{injective}&\text{ if } i = k + 1
\end{cases}. $$
\TODO{Finish proof: $m_k$ well behaved, above assertion.}
\item Suppose $\{v_\alpha\}$ is a basis for $V(0)$. Define $V(0)\to X$ by choosing preimages $x_\alpha$ such that $p(x_\alpha)= g(v_\alpha)$ ($p$ is surjective). Define $\phi(v_\alpha)= x_\alpha$.
\item Suppose $\phi$ has been defined on $V(n)$. Write $V(n+1)= V(n)\oplus V'$ and let $\{v_\alpha\}$ be a basis for $V'$. Then $dv_\alpha\in\Lambda V(n)$, hence $\phi(dv_\alpha)$ is defined and
$$ d \phi d v_\alpha=\phi d^2 v_\alpha=0$$
$$ p \phi d v_\alpha= g d v_\alpha= d g v_\alpha. $$
Now $\phi d v_\alpha$ is a cycle and $p \phi d v_\alpha$ is a boundary of $g v_\alpha$. By the following lemma there is a $x_\alpha\in X$ such that $d x_\alpha=\phi d v_\alpha$ and $p x_\alpha= g v_\alpha$. The former property proves that $\phi$ is a chain map, the latter proves the needed commutativity $p \circ\phi= g$.
\end{itemize}
\end{proof}
\begin{lemma}
Let $p: X \to Y$ be a trivial fibration, $x \in X$ a cycle, $p(x)\in Y$ a boundary of $y' \in Y$. Then there is a $x' \in X$ such that
$$ dx' = x \quad\text{ and }\quad px' = y'. $$
\end{lemma}
\begin{proof}
We have $p^\ast[x]=[px]=0$, since $p^\ast$ is injective we have $x = d \overline{x}$ for some $\overline{x}\in X$. Now $p \overline{x}= y' + db$ for some $b \in Y$. Choose $a \in X$ with $p a = b$, then define $x' =\overline{x}- da$. Now check the requirements: $p x' = p \overline{x}- p a = y'$ and $d x' = d \overline{x}- d d a = d \overline{x}= x$.
\end{proof}
\begin{lemma}
Let $f: X \we Y$ be a weak equivalence between cdga's and $M$ a minimal model for $X$. Then $f$ induces an bijection:
$$ f_\ast: [M, X]\tot{\iso}[M, Y]. $$
\end{lemma}
\begin{proof}
If $f$ is surjective this follows from the fact that $M$ is cofibrant and $f$ being a trivial fibration (see \cite[lemma 4.9]{dwyer}).
In general we can construct a cdga $Z$ and trivial fibrations $X \to Z$ and $Y \to Z$ inducing bijections:
$$[M, X]\tot{\iso}[M, Z]\toti{\iso}[M, Y], $$
compatible with $f_\ast$. \cite[Proposition 12.9]{felix}.
\end{proof}
\begin{lemma}
Let $\phi: (M, d)\we(M', d')$ be a weak equivalence between minimal algebras. Then $\phi$ is an isomorphism.
Let $M$ and $M'$ be generated by $V$ and $V'$. Then $\phi$ induces a weak equivalence on the linear part $\phi_0: V \we V'$\cite[Theorem 1.5.2]{loday}. Since the differentials are decomposable, their linear part vanishes. So we see that $\phi_0: (V, 0)\tot{\iso}(V', 0)$ is an isomorphism.
Conclude that $\phi=\Lambda\phi_0$ is an isomorphism.
\end{proof}
\begin{theorem}
Let $m: (M, d)\we(A, d)$ and $m': (M', d')\we(A, d)$ be two minimal models. Then there is an isomorphism $\phi(M, d)\tot{\iso}(M', d')$ such that $m' \circ\phi\eq m$.
By the previous lemmas we have $[M', M]\iso[M', A]$. By going from right to left we get a map $\phi: M' \to M$ such that $m' \circ\phi\eq m$. On homology we get $H(m')\circ H(\phi)= H(m)$, proving that (2-out-of-3) $\phi$ is a weak equivalence. The previous lemma states that $\phi$ is then an isomorphism.
