Exercise 9.12

The range of $\check{f}$ is a torus with inner radius $b-a$ and outer radius $b+a$, centered at the origin, whose plane is perpendicular to the $z$-axis. To see this, the central circle of such a torus is the set of points $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,0)$. Then the circle of radius $a$ which intersects the half-plane that goes through the $z$-axis and the point $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,0)$ is the circle centered at that point and perpendicular to the $x$- $y$-plane, with radius vector $a(\cos <!--\; FUNCTION\; APPLICATION\; -->t,\sin <!--\; FUNCTION\; APPLICATION\; -->t,0)$, which is described by

Note that $\check{f}$ is a one-to-one mapping of the square $S$ given by $(x,y)$, $0\le x<2\pi$, $0\le y<2\pi$ onto the torus.

(a) Taking the partial derivatives of $f_1$ with respect to $s$ and $t$ to get $f_1$, we want to solve

Restricing ourselves to the square $S$, the first component equals 0 when either $s=0$, $s=\pi$, $t=\pi ∕2$, or $t=3\pi ∕2$. Since $b+a\cos <!--\; FUNCTION\; APPLICATION\; -->s>0$ for all $s$, the second component equals 0 only when $t=0$ or $t=\pi$. Hence $f_1$ equals $\check{0}$ in $S$ only at the four points

which map onto the four points

respectively.

(b) Taking the partial derivatives of $f_3$ with respect to $s$ and $t$ to get $f_3$, we want to solve

which occurs in $S$ at the points $(s,t)$, where $s=\pi ∕2$ or $s=3\pi ∕2$, and $0\le t<2\pi$. The points $(\pi ∕2,t)$, $0\le t<2\pi$, are mapped by $\check{f}$ to the circle $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,a)$, and the points $(3\pi ∕2,t)$, $0\le t<2\pi$, are mapped to the circle $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,-a)$.

(c) At $(b+a,0,0)$, corresponding to $(s,t)=(0,0)$, where $\cos <!--\; FUNCTION\; APPLICATION\; -->s=\cos <!--\; FUNCTION\; APPLICATION\; -->t=1$, $f_1$ attains its maximum value, $b+a$, and at $(-b-a,0,0)$, corresponding to $(s,t)=(0,\pi )$, where $\cos <!--\; FUNCTION\; APPLICATION\; -->s=1$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->t=-1$, $f_1$ attains its minimum value, $-b-a$. At $(b-a,0,0)$, corresponding to $(s,t)=(\pi ,0)$, where $\cos <!--\; FUNCTION\; APPLICATION\; -->s=-1$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->t=1$, $f_1$ increases if you fix $t=0$ and vary $s$, and decreases if you fix $s=\pi$ and vary $t$. Similarly, at $(a-b,0,0$, corresponding to $(s,t)=(0,\pi )$, where $\cos <!--\; FUNCTION\; APPLICATION\; -->s=1$ and $\cos <!--\; FUNCTION\; APPLICATION\; -->t=-1$, $f_1$ decreases if you fix $t=\pi$ and vary $s$, and increases if you fix $s=\pi$ and vary $t$.

The points $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,a)$, $0\le t<2\pi$, correspond to the points $(s,t)=(\pi ∕2,t)$ where $f_3$ attains its maximum value of $a$, and $(b\cos <!--\; FUNCTION\; APPLICATION\; -->t,b\sin <!--\; FUNCTION\; APPLICATION\; -->t,-a)$, $0\le t<2\pi$, correspond to the points $(s,t)=(3\pi ∕2,t)$ where $f_3$ attains its minimum value of $-a$.

(d) To show that the “irrational winding of the torus” (it even has its own Wikipedia page) is dense, I first show a general proposition about continuous functions. Suppose $F$ is a continuous mapping from $X$ to $Y$ (both metric, or even just topological spaces). Then if $D$ is a dense subset of $X$, then $F(D)$ is a dense subset of $F(Y)$. For if not, then there would be a nonempty open set $G$ in $F(Y)$ disjoint from $F(D)$. Then $F^{-1}(D)$ would be an open subset of $X$ disjoint from $D$, contradicting the density of $D$.