Zero differential in Serre spectral sequence for configuration spaces I moved this question from Math StackExchange.
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?
 A: 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.
