I have a question concerning an exercises from a text call Topology and Groupoid authored by Ronald Brown
The question is as follows:
Let $E^2 = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \leq 1\}$. The space $S^1 \times E^2$ is called the solid torus.
Prove that the 3-sphere
$S^3 = \{(x_1, x_2, x_3 , x_4) \in \mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2+(x_4)^2 = 1\}$
is the union of two spaces each homeomorphic to a solid torus and with intersection
homeomorphic to a torus [Consider the subspaces of $S^3$ given by $(x_1)^2 + (x_2)^2 \leq (x_3)^2 + (x_4)^2$ and by $(x_1)^2 + (x_2)^2 \geq (x_3)^2 + (x_4)^2$]
I am not certain I understand the hint from the square bracket in geometric terms.
From what I understand of how the 3-sphere can be constructed, one takes two 2-spheres and superimposes the boundary of one on top of the other and then glues both boundaries together.
The two 2-sphere can be represented as $S^3_+ = \{(x_1, x_2, x_3 , x_4) \in\mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2 = 1, (x_4)^2 \geq 0\}$ and $S^3_- = \{(x_1, x_2, x_3 , x_4) \in\mathbb R^4 : (x_1)^2+(x_2)^2+(x_3)^2 = 1, (x_4)^2 \leq 0\}$
Is the question asking me to show that both $S^3_+$ and $S^3_-$ are individually homeomorphic to the solid torus and $S^3_+ \cap S^3$ is homeomorphic to the torus? If so how does the hint become relevant?
Thanks in advance
Best Answer
Complex numbers make the description a little nicer. The 3-sphere can be represented by the set $\{(z,w) \in \mathbb C^2 : |z|^2+ |w|^2 =2\}$. This contains the subset $|z|=|w|=1$, which is the product of two circles, that is, a torus. This torus is the common boundary of the subsets where $|z|\le 1\le |w|$ and $|w|\le 1\le|z|$. For each of these subsets there is an explicit homeomorphism onto the product of closed disk with a circle (i.e. the solid torus). Namely, $(z,w)\mapsto (z, w/|w|)$ for the first subset, and similarly for the second.