Guillemin and Pollack asks:
Explicitly exhibit enough parameterizations to cover $S^1 \times S^1 \subset R^4$.
My solution is given by $8$ parameterizations (closely following their example of parameterizing a circle):
$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_5(y,w) = (y, \sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$ $$f_6(y,w) = (y, -\sqrt{1 - y^2}, w, \sqrt{1 - w^2})$$ $$f_7(y,w) = (y, \sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$ $$f_8(y,w) = (y, -\sqrt{1 - y^2}, w, -\sqrt{1 - w^2})$$
I believe these are all I would need In terms of parameterizations but I am unsure, can anyone confirm or deny this?
As pointed out by John Hughes and Andrew Hwang I have not considered every parameterization needed and made some mistakes in the existing parameterization. I have included an additional parameterizations and fixed the existing ones:
$$f_1(x,z) = (x, \sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_2(x,z) = (x, -\sqrt{1 - x^2}, z, \sqrt{1 - z^2})$$ $$f_3(x,z) = (x, \sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_4(x,z) = (x, -\sqrt{1 - x^2}, z, -\sqrt{1 - z^2})$$ $$f_5(y,w) = (\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$ $$f_6(y,w) = (-\sqrt{1 - y^2}, y, \sqrt{1 - w^2}, w)$$ $$f_7(y,w) = (\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$ $$f_8(y,w) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - w^2}, w)$$
$$f_9(x,w) = (x, \sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$ $$f_{10}(x,w) = (x, -\sqrt{1 - x^2}, w, \sqrt{1 - w^2})$$ $$f_{11}(x,w) = (x, \sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$ $$f_{12}(x,w) = (x, -\sqrt{1 - x^2}, w, -\sqrt{1 - w^2})$$ $$f_{13}(y,z) = (\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$ $$f_{14}(y,z) = (-\sqrt{1 - y^2}, y, \sqrt{1 - z^2}, z)$$ $$f_{15}(y,z) = (\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$ $$f_{16}(y,z) = (-\sqrt{1 - y^2}, y, -\sqrt{1 - z^2}, z)$$
Note: $f_9$, $f_{10}$ ... are unnecessary, as an appropriate variable substitution can be made.