Fixes a proof (to some extend)

thesis/notes/Minimal_Models.tex

@@ -47,6 +47,8 @@ The above definition is the same as in \cite{felix} without assuming connectivit
 
 It is clear that induction will be an important technique when proving things about (minimal) Sullivan algebras. We will first prove that minimal models always exist for $1$-connected cdga's and afterwards prove uniqueness.
 
+\todo{at the moment this is just cut n pasted. Rewrite to make sense in this context}
+Minimal models admit very nice homotopy groups. Note that for a minimal algebra $\Lambda V$ there is a natural augmentation and the the differential is decomposable. Hence $Q \Lambda V$ is naturally isomorphic to $(V, 0)$. In particular the homotopy groups are simply given by $\pi^n(\Lambda V) = V^n$.
 
 \section{Existence}
 
@@ -128,8 +130,12 @@ Now if the map $f$ is a weak equivalence, both maps $\phi$ and $\psi$ are surjec
 	Let $\phi: (M, d) \we (M', d')$ be a weak equivalence between minimal algebras. Then $\phi$ is an isomorphism.
 \end{lemma}
 \begin{proof}
-	\todo{introduce homotopy groups before this point. Prove it using that} 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.
+	Since both $M$ and $M'$ are minimal, they are cofibrant and so the weak equivalence is a strong homotopy equivalence (\CorollaryRef{cdga_homotopy_properties}). And so the induced map $\pi^n(\phi) : \pi^n(M) \to \pi^n(M')$ is an isomorphism (\LemmaRef{cdga-homotopic-maps-equal-pin}).
+
+	Since $M$ (resp. $M'$) is free as a cga's, it is generated by some graded vector space $V$ (resp. $V'$). By an earlier remark \todo{where?} the homotopy groups were eassy to calculate and we conclude that $\phi$ induces an isomorphism from $V$ to $V'$:
+	\[ \pi^\ast(\phi) : V \tot{\iso} V'. \]
+
+	Conclude that $\phi = \Lambda \phi_0$ \todo{why?} is an isomorphism.
 \end{proof}
 
 \Theorem{unique-minimal-model}{
@@ -155,11 +161,6 @@ Now by \LemmaRef{minimal-model-bijection} we get a bijection ${m_Y}_\ast^{-1} :
 }
 
 
-\section{Homotopy groups of minimal models}
-\todo{at the moment this is just cut n pasted. Rewrite to make sense in this context}
-Minimal models admit very nice homotopy groups. Note that for a minimal algebra $\Lambda V$ there is a natural augmentation and the the differential is decomposable. Hence $Q \Lambda V$ is naturally isomorphic to $(V, 0)$. In particular the homotopy groups are simply given by $\pi^n(\Lambda V) = V^n$.
-
-
 \section{The minimal model of the sphere}
 We know from singular cohomology that the cohomology ring of a $n$-sphere is $\Z[X] / (X^2)$. This allows us to construct a minimal model for $S^n$.
 \Definition{minimal-model-sphere}{