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I am trying to compute homology of $Conf(n, \mathbb{R}^2)$ - ordered configurations of $n$ points on the plane - using Serre spectral sequence. I know that this computations has been done for cohomology in Arnold's *The Cohomology Ring of Colored Braid Group*.


There is one step though which is unclear for me. Namely, he claims that $d\_2$ - the only differential which might be non-zero - is actually zero. 



> 
> The only possible differential $d\_2$ is in fact zero (this easily follows from the existence of the secant of the surface). 
> 
> 
> 


Why this is true?



Zero differential in Serre spectral sequence for configuration spaces
I'll write $C\_n$ for the configuration space, and $X\_n$ for $\mathbb{R}^2$ with $n$ points removed. You are presumably thinking about the spectral sequence
$$ E\_2^{pq} = H^p(C\_{n-1};H^q(X\_{n-1})) \Longrightarrow H^{p+q}(C\_n), $$
with differentials $d\_r\colon E\_r^{pq}\to E\_r^{p+r,q+1-r}$. It is a key point that the projection $\pi\colon C\_n\to C\_{n-1}$ has a section: for example, we can use
$$ \sigma(z\_1,\dotsc,z\_{n-1}) = 
 \left(z\_1,\dotsc,z\_{n-1},(\max(\|z\_1\|,\dotsc,\|z\_{n-1}\|)+1).(1,0)\right).
$$
This implies that the map $\pi\_1(C\_n)\to\pi\_1(C\_{n-1})$ is surjective, and thus that $\pi\_1(C\_{n-1})$ acts trivially on $H^\*(X\_{n-1})$. This is also a free abelian group by induction on $n$, so the $E\_2$ term can be rewritten as $H^p(C\_{n-1})\otimes H^q(X\_{n-1})$. It is clear that $H^\*(X\_{n-1})$ is generated by classes in degree $1$, and it will suffice to show that these support no differentials. Any differentials after $d\_2$ land in the lower half plane and so are zero. The differential $d\_2$ on $H^1(X\_{n-1})$ lands in $H^2(C\_{n-1})$, and anything in the image will be killed by the map $H^2(C\_{n-1})\to H^2(C\_n)$ (by the basic definition and properties of the Serre spectral sequence). However, that map is injective, because of the existence of the section $\sigma$. Thus, $d\_2$ is zero and the spectral sequence collapses.


