I am attempting to compute the (integral) cohomology ring structure of the 3 configuration space of $\mathbb{R}^m$ and have run into a few doubts.
Using a result of Fadell and Neuwirth, we have that conf($\mathbb{R}^m$, 3) is the total space of the fibration:
p: conf($\mathbb{R}^m$, 3) $\rightarrow$ conf($\mathbb{R}^m$, 2) given by the obvious map onto the first 2 factors with fibre homeomorphic to $\mathbb{R}^m - Q$ where $Q$ is two points.
Since $\mathbb{R}^m - Q $ is homotopy equivalent to a wedge of two $m-1$ spheres, it has cohomology groups $\mathbb{Z}$ and $\mathbb{Z}\bigoplus \mathbb{Z}$ in dimensions $0$ and $m-1$ respectfully (Mayer-Vietoris).
Similarly conf($\mathbb{R}^m$, 2) is homotopy equivalent to $S^{m-1}$ where a map is $(x,y) \rightarrow \frac{(x-y)}{|x-y]}$. Assuming trival local coefficients (where the only problem could arise when $m=2$ since otherwise the base is simply connected) we can apply the Serre spectral sequence with $E_2^{p,q} = H^p(S^{m-1}, H^q(\vee_2 S^{m-1} )) $.
An application of the UCT (since everything is free) gives that $E_2^{p,q} = H^p(S^{m-1}) \otimes H^q(\vee_2 S^{m-1} ) $
We then have that $E_2 = E_{\infty}$ since we only have cohomology in degree $0$ and $m-1$ for the base and fibre and hence there can be no non-trivial differentials.
This leads me to a few questions.
In general the $E_{\infty}$ product structure doesn't determine the product structure on the total space (see for example p 29 SSAT ) but things work out in this instance since everything is a free $\mathbb{Z}$ module and of finite type? What is a necessary and sufficient condition on the $E_{\infty}$ page structure to guarantee that the product structures coincide?
I know that $H^*(S^{m-1}) = \mathbb{Z}[a_1]/(a_1^2)$ where $|a_1| = m-1$ and that $H^*(S^{m-1} \vee S^{m-1}) = \mathbb{Z}[a_2]/(a_2^2) \times \mathbb{Z}[a_3]/(a_3^2)$ where $|a_2|=|a_3| = m-1$.
Does this simply imply that $H^*($conf($\mathbb{R}^m$, 3) ) $\simeq \mathbb{Z}[a_1]/(a_1^2)$ $\otimes \mathbb{Z}[a_2]/(a_2^2) \times \mathbb{Z}[a_3]/(a_3^2)\simeq \mathbb{Z}[a_1,a_2,a_3]/(a_1^2=a_2^2=a_3^2)$? Am I missing something?