
#1
Aug911, 11:04 AM

P: 429

Hello all.
For some reason or another, I've been thinking about the following space: Take two copies of the square. Identify their edges as you would usually to form a torus, but identify them also to the corresponding edges of the counterpart sqaure (so we have two tori, joined in some way). Maybe I should think a bit harder about what this space is, but am I right in saying that the universal cover of this space is simply a "plane of spheres"? Indeed, if it was just the one square (with identifications) the universal cover would be the plane, so it seems to me that the universal cover of this one is two planes, but where we associate the grid lines, so it might be easier imagine a plane of octahedrons. Is what I have said correct? Would the homotopy of this space be pi_1=trivial (must be for universal cover) pi_{2}=ZxZxZxZx....... pi_{n}=0 for n>2? (so this is a K(ZxZxZx...,2)) I was hoping the universal cover of this space would be trivial you see :( 



#2
Aug911, 11:12 AM

P: 429

Actually, I'm guessing that my pi_2 could be far more complicated.




#3
Aug1211, 07:22 AM

P: 429

Here is another question:
Suppose I have a covering space C of a topological space X (we'll assume that X is a CW complex). If C is path connected and the fibre of the projection map is infinite, must C be the universal cover of X? I can't think of any counterexamples to this. 



#4
Aug1211, 08:20 PM

P: 491

Funny Universal Cover
Doesn't R x S^1 cover the torus?




#5
Aug1411, 06:00 AM

P: 429

Sorry, that was a really dumb question given a space X with fundamental group G, for any subgroup H<G there exists a cover of X with fundamental group H, of course including H infinite :/
So yes, the cylinder is a good example of such a cover. 


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