I was trying to understand this problem:
Construct a simply-connected covering space of the space $X \subset \mathbb R^3$ that is the union of a sphere and a diameter.
And my idea was to only use two spheres separated by a line, but when I looked here Covering spaces need big help Hatcher I found that there has to be infinite number of spheres, is this because we want a universal cover because we want a simply connected cover and also because we want to have loops (start and end at the same time)? Would it break the homomorphism condition because it would not be locally homeomorphic ? If so, why the local homeomorphic condition will be broken?
Any explanations will be greatly appreciated!
Best Answer
Let me use two spheres then. Image the line $[-1,1]\times\{(0,0)\}$ embedded in $\Bbb R^3$, the unit spheres positioned with centres $(-2,0,0)$ and $(2,0,0)$. Use the obvious kind of projection map from here onto $X$ which is the union of the same line with the unit sphere centred at the origin.
Consider the part of the sphere at $(3,0,0)$. This point maps to $(-1,0,0)$. Any neighbourhood of that point in $X$ catches some small segment of the line; however the preimage of that neighbourhood in our "covering" space is just some sphere-neighbourhood of $(3,0,0)$ with no line presented. Local homeomorphy is lost. You need to attach a further line from $3\to 5$ to make this work. Oh, but then you need to attach a sphere which touches the line at $(5,0,0)$ for the same sorts of reason. Oh but then you have to attached another line. And so on...