homotopy type of complement of subspace arrangement I am studying the homotopy type of a space,and i hope it would be a $K(\pi,1)$ space.
now i have find its covering,once we can say the covering is $K(\pi,1)$,so is the space
itself.and the covering is 
$\mathbb{R}^4-M$ where $M=M_1\cup M_2\cup M_3\cup M_4$,
$M_1=\{(x,y,z,w)|x,y \in \mathbb R,z,w \in\mathbb Z\}$
$M_2=\{(x,y,z,w)|y,z \in \mathbb R,x,w \in\mathbb Z\}$
$M_3=\{(x,y,z,w)|x,w \in \mathbb R,y,z \in\mathbb Z\}$
$M_4=\{(x,y,z,w)|z,w \in \mathbb R,x,y \in\mathbb Z\}$
I guess $\mathbb{R}^4-M$ is $K(\pi,1)$ space,can someone help prove this?
 A: Since you arrangement is so simple, there are must be some purely topological methods to check asphericity (or non-asphericity) of its complement. This is a well-developed subject I do not know much about. 
I am more familiar with the geometric approach developed by Daniel Allcock (following predictions of Gromov) in  this paper. Unfortunately, the result in Allcock's does not apply to your arrangement because it is not "normal" (as defined in the paper). Allcock has a sequel in works where he gives a definitive answer to the question of asphericity for a large class of arrangements including the one you care about; if everything fails you might want to ask him for a preliminary version. 
A: Here is a sketch of a solution. As Petya tells us, $\mathbf{R}^4\setminus M$ is invariant under the action of $\mathbf{Z}^4$ by integral translations, so it suffices to show that the image of $\mathbf{R}^4\setminus M$ in $T^4=\mathbf{R}^4/T^4$ is a $K(\pi,1)$. Denote this image by $N$. It is mapped to the 2-torus by $p:[x,y,z,w]\mapsto [x,y]$ where the square brackets stand for the images in the tori. $N$ is homotopy equivalent to $p^{-1}$ of the wedge $W$ of two circles given by $[x]=[1/2]$ and $[y]=[1/2]$. Set $X=S^1\times (S^1\vee S^1)$. The preimage $p^{-1}(W)$ is homotopy equivalent to two copies of $X$ joined along $pt\times (S^1\vee S^1)$, with a circle in each of the copies deleted; the deleted circles correspond to the two different circles of $S^1\vee S^1$, so $p^{-1}(W)$ is homotopy equivalent to the wedge of two 2-tori, and hence so is $N$.
