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algori
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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

$R^4-M$$\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 $R^4-M$$\mathbb{R}^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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

$R^4-M$ 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 $R^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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?

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François G. Dorais
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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

$R^4-M$where 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 $R^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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

$R^4-M$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 $R^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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

$R^4-M$ 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 $R^4-M$ is $K(\pi,1)$ space,can someone help prove this?

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student
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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

$R^4-M$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 $R^4-M$ is $K(\pi,1)$ space,can someone help prove this?