I've been reading Galatius's Park City notes on the Madsen-Weiss theorem (available [here](http://math.stanford.edu/~galatius/notes2.pdf)).


On page 8, he states the following theorem. Let $X$ be a space such that $\pi\_1(X)$ is abelian and acts trivially on the rational cohomology of the universal cover of $X$. Let $V = \oplus\_{n \geq 1} V\_n$ be a graded vector space and $V \rightarrow H^{\ast}(X;\mathbb{Q})$ a linear map such that the induced map $\mathbb{Q}[V] \rightarrow H^{\ast}(X;\mathbb{Q})$ is an isomorphism in degrees $\leq n$. Then there is a map $\mathbb{Q}[s^{-1} V] \rightarrow H^{\ast}(\Omega X;\mathbb{Q})$ which is an isomorphism in degrees $\leq n-1$. Here $s^{-1} V$ is the result of shifting the gradings on $V$ down by one.


He claims that this can be proved by "This can be proved using the Serre spectral sequence for the path-loop fibration over the universal cover $\tilde{X}$".


Despite a lot of work, I have been unable to fill in the details of this use of the Serre spectral sequence. It seems weird -- how on earth is the homology of the universal cover at all related to the homology of $X$?


Can anyone help me? Thank you very much.



Homology of loop space
I believe the argument Galatius had in mind is the following. Let us write $\bar{V} = \oplus\_{n \geq 2} V\_n$, so $V = V\_1 \oplus \bar{V}$. All cohomology will be rational, and we write $G := \pi\_1(X)$. Lets go ahead and suppose $G$ is finitely-generated and $X$ has rational cohomology of finite type.


By assumption $V\_1 \to H^1(X) = H^1(G)$ is an isomorphism. As $G$ is an abelian group the extension to a free commutative graded algebra $\mathbb{Q}[V\_1] \to H^\*(G)$ is an isomorphism. We also have the composition $\mathbb{Q}[\bar{V}] \hookrightarrow H^\*(X) \to H^\*(\widetilde{X})$, and I claim this is an isomorphism in degrees $\leq n$.


To see this, we apply the spectral sequence described by Emerton in his answer, which goes
$$E\_2^{p,q} = H^p(G ; H^q(\widetilde{X})) \Longrightarrow H^{p+q}(X).$$
As the action of $G$ on $H^q(\widetilde{X})$ is trivial, we can re-write $E\_2$ as $H^\*(G) \otimes H^\*(\widetilde{X}) = \mathbb{Q}[V\_1] \otimes H^\*(\widetilde{X})$. The image of $\mathbb{Q}[\bar{V}] \hookrightarrow H^\*(X) \to H^\*(\widetilde{X})$ can support no differentials (as these classes are pulled back from $X$, by construction). The map $\mathbb{Q}[\bar{V}] \hookrightarrow H^\*(X) \to H^\*(\widetilde{X})$ is an isomorphism in degree 0 and 1 (this is obvious, as there is nothing in degree 1 on both sides). In the minimal degree in which it is not as isomorphism (below $n$), we see we get a contradiction. If it is not surjective, there are extra classes in $H^\*(\widetilde{X})$. These must support a differential, but we know everything in lower degrees is correct, so there is nothing for such a differential to hit. Otherwise, the map is surjective but not injective. But then we do not find enough cohomology in this degree to produce the result on $E\_\infty$, which we know to be $\mathbb{Q}[V]$.


So, we have that $\mathbb{Q}[\bar{V}] \to H^\*(\widetilde{X})$ is an isomorphism in degrees $\leq n$. Now we use the path fibration
$$\Omega\_\bullet X \simeq \Omega \widetilde{X} \longrightarrow P\widetilde{X} \longrightarrow \widetilde{X},$$
where the fibre is identified with the basepoint component of the loop space of $X$. We are now in the situation of (1) in Goodwillie's answer. We have the map $c: \mathbb{Q}[s^{-1} \bar{V}] \to H^\*(\Omega\_\bullet X)$ and want to see that it is an isomorphism. In low degrees the first differential is
$$d\_2 : H^1(\Omega\_\bullet X) \longrightarrow H^2(\widetilde{X}) = V\_2$$
and this must be an isomorphism as the spectral sequence converges to zero. This shows that $c$ is an isomorphism in degree 1. Now we argue in a similar way as above by thinking about the minimal degree in which it is not an isomorphism: this argument takes us up to degree $n-1$.


