@ -188,7 +188,7 @@ In particular if the vector space $V'$ is finitely generated, we can repeat this
Now $V(n)$ is finitely generated for all $n$ by assumption. By the inductive procedure above we see that $(\Lambda V(n), d)\to A(K(\Lambda V(n), d))$ is a weak equivalence for all $n$. Hence $(\Lambda V, d)\to A(K(\Lambda V, d))$ is a weak equivalence.
Now $V(n)$ is finitely generated for all $n$ by assumption. By the inductive procedure above we see that $(\Lambda V(n), d)\to A(K(\Lambda V(n), d))$ is a weak equivalence for all $n$. Hence $(\Lambda V, d)\to A(K(\Lambda V, d))$ is a weak equivalence.
}
}
Now we want to prove that $X \to K(M(X))$ is a weak equivalence for a simply connected rational space $X$ of finite type. For this, we will use that $A$ preserves and detects such weak equivalences by \CorollaryRef{serre-whitehead} (the Serre-Whitehead theorem). To be precise: $X \to K(M(X))$ is a weak equivalence if and only if $A(K(M(X)))\to A(X)$ is a weak equivalence.
Now we want to prove that $X \to K(M(X))$ is a weak equivalence for a simply connected rational space $X$ of finite type. For this, we will use that $A$ preserves and detects such weak equivalences by \CorollaryRef{serre-whitehead} (the Serre-Whitehead theorem). To be precise: for a simply connected rational space $X$ the map $X \to K(M(X))$ is a weak equivalence if and only if $A(K(M(X)))\to A(X)$ is a weak equivalence.
\Lemma{}{
\Lemma{}{
The map $X \to K(M(X))$ is a weak equivalence for simply connected rational spaces $X$ of finite type.
The map $X \to K(M(X))$ is a weak equivalence for simply connected rational spaces $X$ of finite type.
@ -7,7 +7,6 @@ Given a category $\cat{C}$ and a functor $F: \DELTA \to \cat{C}$, then define th
F^\ast(C)_n &= \Hom_{\cat{C}}(F[n], Y) & C \in\cat{C}
F^\ast(C)_n &= \Hom_{\cat{C}}(F[n], Y) & C \in\cat{C}
\end{align*}
\end{align*}
A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimits. Applying $F$ on these diagrams, make it clear that $F_!$ is functorial. Secondly we see readily that $F^\ast$ is functorial. By using the definition of colimit and the Yoneda lemma (Y) we can prove that $F_!$ is left adjoint to $F^\ast$:
A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimits. Applying $F$ on these diagrams, make it clear that $F_!$ is functorial. Secondly we see readily that $F^\ast$ is functorial. By using the definition of colimit and the Yoneda lemma (Y) we can prove that $F_!$ is left adjoint to $F^\ast$:
@ -19,18 +18,14 @@ A simplicial map $X \to Y$ induces a map of the diagrams of which we take colimi
\end{align*}
\end{align*}
Furthermore we have $F_!\circ\Delta[-]\iso F$. In short we have the following:
Furthermore we have $F_!\circ\Delta[-]\iso F$. In short we have the following:
\cdiagram{Kan_Extension}
\cdiagram{Kan_Extension}
In our case where $F =\Apl$ and $\cat{C}=\CDGA_\k$ we get:
In our case where $F =\Apl$ and $\cat{C}=\CDGA_\k$ we get:
\cdiagram{Apl_Extension}
\cdiagram{Apl_Extension}
\subsection{The cochain complex of polynomial forms}
\subsection{The cochain complex of polynomial forms}
In our case we take the opposite category, so the definition of $A$ is in terms of a limit instead of colimit. This allows us to give a nicer description:
In our case we take the opposite category, so the definition of $A$ is in terms of a limit instead of colimit. This allows us to give a nicer description:
\begin{align*}
\begin{align*}
A(X)
A(X)
&= \lim_{\Delta[n]\to X}\Apl_n
&= \lim_{\Delta[n]\to X}\Apl_n
@ -38,9 +33,7 @@ In our case we take the opposite category, so the definition of $A$ is in terms
where the addition, multiplication and differential are defined pointwise. Conclude that we have the following contravariant functors (which form an adjoint pair):
where the addition, multiplication and differential are defined pointwise. Conclude that we have the following contravariant functors (which form an adjoint pair):
\begin{align*}
\begin{align*}
A(X) &= \Hom_\sSet(X, \Apl) & X \in\sSet\\
A(X) &= \Hom_\sSet(X, \Apl) & X \in\sSet\\
K(C)_n &= \Hom_{\CDGA_\k}(C, \Apl_n) & C \in\CDGA_\k.
K(C)_n &= \Hom_{\CDGA_\k}(C, \Apl_n) & C \in\CDGA_\k.
@ -52,7 +45,6 @@ where the addition, multiplication and differential are defined pointwise. Concl
Another way to model the $n$-simplex is by the singular cochain complex associated to the topological $n$-simplices. Define the following (non-commutative) dga's \todo{Choose: normalized or not?}:
Another way to model the $n$-simplex is by the singular cochain complex associated to the topological $n$-simplices. Define the following (non-commutative) dga's \todo{Choose: normalized or not?}:
$$ C_n = C^\ast(\Delta^n; \k). $$
$$ C_n = C^\ast(\Delta^n; \k). $$
The inclusion maps $d^i : \Delta^n \to\Delta^{n+1}$ and the maps $s^i: \Delta^n \to\Delta^{n-1}$ induce face and degeneracy maps on the dga's $C_n$, turning $C$ into a simplicial dga. Again we can extend this to functors by Kan extensions
The inclusion maps $d^i : \Delta^n \to\Delta^{n+1}$ and the maps $s^i: \Delta^n \to\Delta^{n-1}$ induce face and degeneracy maps on the dga's $C_n$, turning $C$ into a simplicial dga. Again we can extend this to functors by Kan extensions
\cdiagram{C_Extension}
\cdiagram{C_Extension}
where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix}. We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix}. We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
@ -148,4 +140,3 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
So by the five lemma we can conclude that the middle morphism is an isomorphism as well, proving the isomorphism $H^n(A(X))\tot{\iso} H^n(C^\ast(X))$ for all $n$. This proves the statement for all $X$.
So by the five lemma we can conclude that the middle morphism is an isomorphism as well, proving the isomorphism $H^n(A(X))\tot{\iso} H^n(C^\ast(X))$ for all $n$. This proves the statement for all $X$.
Joshua Moerman
10 years ago