@ -39,24 +39,20 @@ The requirement that there exists a filtration can be replaced by a stronger sta
\section{Existence}
\begin{theorem}
Let $(A, d)$ be an $0$-connected cdga, then it has a Sullivan model $(\Lambda V, d)$. Furthermore if $(A, d)$ is $r$-connected,$V^i =0$ for all $i \leq r$.
Let $(A, d)$ be an $0$-connected cdga, then it has a Sullivan model $(\Lambda V, d)$. Furthermore if $(A, d)$ is $r$-connected with $r \geq1$ then$V^i =0$ for all $i \leq r$ and in particular $(\Lambda V, d)$ is minimal.
\end{theorem}
\begin{proof}
Start by setting $V(0)= H^{\geq1}(A)$ and $d =0$. This extends to a morphism $m_0 : (\Lambda V(0), 0)\to(A, d)$.
Note that the freeness introduces products such that the map $H(m_0) : H(\Lambda V(0))\to H(A)$ is not an isomorphism. We will ``kill'' these defects inductively.
Note that the freeness introduces products such that the map $H(m_0) : H(\Lambda V(0))\to H(A)$ is \emph{not} an isomorphism. We will ``kill'' these defects inductively.
Suppose $V(k)$ and $m_k$ are constructed. Consider the defect $\ker H(m_k)$ and let $\{[z_\alpha]\}_{\alpha\in A}$ be a basis for it. Define $V_{k+1}=\bigoplus_{\alpha\in A}\k\cdot v_\alpha$ with the degrees $\deg{v_\alpha}=\deg{z_\alpha}-1$.
Now extend the differential by defining $d(v_\alpha)= z_\alpha$. This step kills the defect, but also introduces new defects which will be killed later. Notice that $z_\alpha$ is a cocycle and hence $d^2 v_\alpha=0$.
Since $[z_\alpha]$ is in the kernel of $H(m_k)$ we see that $m_k z_\alpha= d a_\alpha$ for some $a_\alpha$. Extend $m_k$ to $m_{k+1}$ by defining $m_{k+1}(v_\alpha)= a_\alpha$. Notice that $m_{k+1} d v_\alpha= m_{k+1} z_\alpha= d a_\alpha= d m_{k+1} v_\alpha$.
Now take $V(k+1)= V(k)\oplus V_{k+1}$. We already proved that $d$ is indeed a differential, and that $m_{k+1}$ is indeed a chain map.
Now extend the differential by defining $d(v_\alpha)= z_\alpha$. This step kills the defect, but also introduces new defects which will be killed later. Notice that $z_\alpha$ is a cocycle and hence $d^2 v_\alpha=0$, so $d$ is still a differential.
Since $[z_\alpha]$ is in the kernel of $H(m_k)$ we see that $m_k z_\alpha= d a_\alpha$ for some $a_\alpha$. Extend $m_k$ to $m_{k+1}$ by defining $m_{k+1}(v_\alpha)= a_\alpha$. Notice that $m_{k+1} d v_\alpha= m_{k+1} z_\alpha= d a_\alpha= d m_{k+1} v_\alpha$, so $m_{k+1}$ is a cochain map.
Now take $V(k+1)= V(k)\oplus V_{k+1}$.
Complete the construction by taking the union: $V =\bigcup_k V(k)$. Clearly $H(m)$ is surjective, this was establsihed in the first step. Now if $H(m)[z]=0$, then we know $z \in\Lambda V(k)$ for some stage $k$ and hence by construction is was killed, i.e. $[z]=0$. So we see that $m$ is a quasi isomorphism and by construction $(\Lambda V, d)$ is a sullivan algebra.
\todo{minimality for $1$-connected}
Now assume that $(A, d)$ is $r$-connected ($r \geq1$), this means that $H^i(A)=0$ for all $1\leq i \leq r$, and so $V(0)^i =0$ for all $i \leq r$. Now $H(m_0)$ is injective on $\Lambda^{\leq1} V(0)$, and so the defects are in $\Lambda^{\geq2} V(0)$ and have at least degree $2(r+1)$. This means two things in the first inductive step of the construction. First, the newly added elements have decomposable differential. Secondly, these elements are at least of degree $2(r+1)-1$. After adding these elements, the new defects are in $\Lambda^{\geq2} V(1)$ and have at least degree $2(2(r+1)-1)$. We see that as the construction continues, the degrees of adjoined elements go up. Hence $V^i =0$ for all $i \leq r$ and by the previous lemma $(\Lambda V, d)$ is minimal.
\end{proof}
@ -65,14 +61,14 @@ The requirement that there exists a filtration can be replaced by a stronger sta
Before we state the uniqueness theorem we need some more properties of minimal models.
\begin{lemma}
Sullivan algebras are cofibrant.
Sullivan algebras are cofibrant and the inclusions $(\Lambda V(k), d)\to(\Lambda V(k+1), d)$ are cofibrations.
\end{lemma}
\begin{proof}
Consider the following lifting problem, where $\Lambda V$ is a Sullivan algebra.
\cimage[scale=0.5]{Sullivan_Lifting}
By the left adjointness of $\Lambda$ we only have to specify a map $\phi: V \to X$ such that $p \circ\phi= g$. We will do this by induction.
By the left adjointness of $\Lambda$ we only have to specify a map $\phi: V \to X$ such that $p \circ\phi= g$. We will do this by induction. Note that the induction step proves precisely that $(\Lambda V(k), d)\to(\Lambda V(k+1), d)$ is a cofibrations.
\begin{itemize}
\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
@ -90,16 +86,18 @@ Before we state the uniqueness theorem we need some more properties of minimal m
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:
\Lemma{minimal-model-bijection}{
Let $f: X \we Y$ be a weak equivalence between cdga's and $M$ a minimal algebra. 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}).
If $f$ is surjective this follows from the fact that $M$ is cofibrant and $f$ being a trivial fibration, see \CorollaryRef{cdga_homotopy_properties}.
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}.
\todo{Put this surjectivity trick in a lemma} In general we will reduce to the surjective case. Let $C$ be any cochain complex and define $\delta C^k = C^{k-1}$. Now $(C \oplus\delta C, \delta)$ is again a cochain complex and there is a surjective map to $C$. Define $E(Y)=\Lambda(Y \oplus\delta Y, \delta)$ (we consider $Y$ as a cochain complex). We obtain:
$$ f: X \tot{x \mapsto x \tensor1} X \tensor E(Y)\tot{x \tensor y \mapsto f(x)\cdot y} Y. $$
Now the first map has a left inverse. We have two trivial fibrations $E(Y)\to X$ and $E(Y)\to Y$. This induces
$$[M, X]\toti{\iso}[M, E(Y)]\tot{\iso}[M, Y], $$
compatible with $f_\ast$.
\end{proof}
\begin{lemma}
@ -117,6 +115,20 @@ Before we state the uniqueness theorem we need some more properties of minimal 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.
\end{proof}
The assignment of $X$ to its minimal model $M_X =(\Lambda V, d)$ can be extended to morphisms. Let $X$ and $Y$ be two cdga's and $f: X \to Y$ be a map. By considering their minimal models we get the following diagram.
Now by \LemmaRef{minimal-model-bijection} we get a bijection ${m_Y}_\ast^{-1} : [M_X, Y]\iso[M_X, M_Y]$. This gives a map $M(f)={m_Y}_\ast^{-1}(f m_X)$ from $M_X$ to $M_Y$. Of course this does not define a functor of cdga's as it is only well defined on homotopy classes. However it is clear that it does define a functor on the homotopy category of cdga's.
\Corollary{}{
The assignment $X \mapsto M_X$ defines a functor $M: \Ho(\CDGA^1_\Q)\to\Ho(\CDGA^1_\Q)$. Moreover, since the minimal model is weakly equivalent, $M$ gives an equivalence of categories:
Joshua Moerman
10 years ago