Does anyone know a nice space that this is homeomorphic to? Its

6 disjoint copies of R^2 tb.h

>>8150641
The surface of a cube minus the edges, yes. This is the same as >>8150581

>>8150581
Yes. If OP wants a formal name for it for some reason, it's the 6-fold coproduct [math]\coprod_{i=1}^6 \mathbb{R}^2 [/math].

>>8150581
>>8150641
>>8150674

Does it not matter whatsoever that each face of the cube is adjacent to four others, while in our original space each face is adjacent to only two others?

>>8150728
Topologically it doesn't matter.

>>8150728
Topologically the faces may as well be a million miles apart.

>>8150581
I was actually hoping for something nicer. An example would be if we had R^2 take away x=y and x=-y that would be homeomorphic to (R^*)^2 where R^* is just R-{0}.

The reason for the is that Im trying to find out, in C^3 what the space C^3 take away the hyperplanes described by x-y=0, x-z=0 and y-z=0 is homeomorphic to. And I see that these hyper planes can be arranged so that this can become a "4" dimensional problem instead of a "6". And by making it into a 4 dimensional problem this is the type of graph I get. So I'd really want an answer of the form C-{blah}\times C-{blah} \times C.

>>8150511
Homeomorphic means a space is "equivalent" to a deformed space. Example a surface of an ellipsoid is homeomorphic to the surface of a sphere.

With this in mind the universe subject to general relativity, which distorts space locally, is homeomorphic to your linear space.

>>8151353
Isn't it just (R^*)^6?

>>8151438
I guess it is but I was hoping for something cleaner, because then every answer would be just R^n depending on what hyperplanes I take out.