In completing an exercise I have shown that $\Bbb{R}^3$ minus the axes $x=0$, $y=0$, and $z=0$ is homotopic equivalent to the cube graph $Q_3$. To visualize this, $\Bbb{R}^3-0$ is homotopy equivalent to $S^2$; then, remove the $x=0$ axis and we have a space equivalent to "a cylinder"; finally,remove the $y=0$ and $x=0$ axes and we have a punctured cylinder: expanding the punctures gives the graph.
The next part of the question asks me to consider $\Bbb{R}^4$ minus the planes $x=y=0$, $z=w=0$. I tried a similar approach, however $\Bbb{R}^4$ is much more difficult to imagine. Start with $\Bbb{R}^4-0\simeq S^3$. I think that removing the $x=y=0$ axis gives a surface in $\Bbb{R}^4$ somewhat like a cylinder but "enclosing" the $x=y=0$ plane. The next cut/puncture I have no idea with, however. I cannot think how to remove $z=w=0$, without leaving the resulting surface disconnected.
Another idea I had was by trying to construct something analogous to $Q_3$. Where $Q_3$ had eight vertices, one corresponding to each octant, perhaps my surface in $\Bbb{R}^4$ will have sixteen edges with thirty-two surfaces corresponding to the adjacent orthants (sharing a hyperplane).
