This map is a covering map by the same argument from Theorem 2.
p on its own takes each unit interval in ℝ and wraps it around the unit circle. p×p takes each unit square
[n,n+1]×[m,m+1] where n,m∈ℝ and wraps it around the torus.
In reality this would be a surface in ℝ4, but to make it easier to visualize we have dropped down to to ℝ3
Let’s look at a specific example. Let D be the donut-shaped surface that is obtained from rotating the circle C1 centered at (1,0,0) and with radius 13 in the xz plane about the z axis.
Let C2 be the circle of radius 1 centered at the origin. So we have an image that looks like this:
We can show that S1×S1 is homeomorphic to D. Let a be a point on the circle C1 and let b be a point on C2. Define
f:C1×C2⟶Dby
defining f(a×b) to be the the point where a ends up after rotating C1 around the z axis until its center hits point b.
We can show that this is homeomorphism of C1×C2 with D. Firstly, we can see that f is continuous since it involves only rotating continuously about the z axis. The inverse will also be continuous since it simply rotates backwards.
To show that it f is bijective, we will employ two different approaches. First, we will show that f is surjective.
Let x0 be any point that lies on our torus lite. We can draw a circle which lies on the torus that contains it. This circle has a center b0. Furthermore, we can measure some angle from the axis of rotation. That angle, when places on the original C1, will correspond to a. Therefore, every point is mapped to, and f is surjective.
Now, to show that f is injective, we can come up with the formula that for f(a,b)
For clarity, we can first express C1 and C2 in Cartesian coordinates
C2=x2+y2=1
C1=(x−1)2+z2=19
Now we can see that the formula
f(a×b)
b=(cosφ,sinφ,0)0≤φ≤2π
gives us any point on C2 and
a=(13cosθ+1,0,13sinθ)0≤θ≤2π
gives us any point on C1. The x component will correspond to the radius of the partial circle drawn by f(a×b) while the z terms corresponds to the height of that partial circle
This gives us the map
f(a×b)=((13cosθ+1)cosφ,(13cosθ+1)sinφ,13sinθ)
For 0≤θ,φ≤2π…
If
(13cosθ+1)cosφ=(13cosθ′+1)cosφ′
(13cosθ+1)sinφ=(13cosθ′+1)sinφ′
13sinθ=13sinθ′
only if θ=θ′ and φ=φ′
Just looking at the first equation, assume that θ≠θ′. In order for the first equality to be true, we need either
θ=π4,θ′=7π4orviceversa, or
θ=3π4,θ′=5π4
But if we look at the third equality, both of these pairs will give us
13sinθ=−13sinθ′
Hence, there is no way for all three equalities to hold unless we have θ=θ′ and, using the same reasoning, φ=φ′, proving injectivity.
We will end this section with a few more examples of covering maps which involve Cartesian products.
Example 3.11
Consider the covering map
p×p:ℝ×ℝ⟶S1×S1
Let b0 be the point p(0)∈S1 and let B0 be the subspace
B0=(S1×b0)∪(S1×b0)
In other words, B0 is the union of two circles that have a single point b0 in common. We call this the figure eight space .
The space
E0=p−1(B0)
is the infinite grid
E0=(ℝ×ℤ)×(ℝ×ℤ)
Thus, the map
P0:E0⟶B0
obtained by restricting p×p is a covering map.
Example 3.12
Consider the covering map
p×i:ℝ×ℝ+⟶S1×ℝ+
where i is the identity map on ℝ+ and p is the map
p(x)=(cosx,sinx)
Take the standard homeomorphism of S1×ℝ+ with ℝ2/{0}. That is,
f((cosx,sinx),r)=(rcosx,rsinx)
This gives us a covering map
ℝ×ℝ+⟶ℝ2/{0}
