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The following exercise is given in the book Topology- James Munkres (2nd Edition, ch-9, sec-54). I have some difficulties in solving this one. I try to describe my thoughts here:

Exercise 5. Consider the path
$$f(t) = (cos 2pi t, sin 2pi t) times (cos 4pi t, sin 4pi t)$$
in $S^1times S^1$. Sketch what $f$ looks like when $S^1times S^1$ is identified with the doughnut surface $D$.

Here, doughnut-shaped surface $D$ in $BbbR^3$ is obtained by rotating the circle $C_1$ in the $xz$-plane of radius $frac13$ centered at $(1, 0,0)$ about the $z$-axis.

My Attempt. Here, $f:[0,1]to S^1times S^1$. First I try to find out a homeomorphism from $S^1times S^1$ to $D$. Well, Munkres gave an example of such homeomorphism, $h$, say:

Consider $C_2$ be the circle of radius $1$ in the $xy$-plane centered at origin. Then we can map $C_1 times C_2$ into $D$ by defining $h(a,b)$ to be the point into which $a$ is carried when one rotates the circle $C_1$ about the $z$-axis until its center hits the point $b$.

Here, $S^1times S^1$ is homeomorphic with $C_1 times C_2$.

Therefore $h circ f:[0, 1]to D$ is the corresponding path in $D$.
So, now everything depends on $h$. I can visualize $h$ as given in its description above. But I need its analytical description also so that I can compute $h(f(t))$.

Let $a=(a_1,0,a_3)in C_1$ and $b=(b_1, b_2, 0)in C_2$ so that $(a,b) in C_1 times C_2$. I have to find the expression for $h(a,b)=(h_1(a,b), h_2(a,b), h_3(a,b))in D$.

But I cannot proceed further. Any hints to find out $h_1, h_2, h_3$ or any better way to solve this is appreciated. Thank you.

The circle in plane $xz$ that is rotated to obtain surface $D$ has a parametrization:
$$ C_1 = leftleft(Big(1+frac13costhetaBig),0,frac13sinthetaright): thetain[0,2pi)right$$
As per the description the homeomorphism $h$, you take a point of $C_1$ an rotate it appropriately around the $z$-axis:
beginalign hbig((costheta,sintheta)times(cosphi,sinphi)big) &= hatR_z(phi)left(Big(1+frac13costhetaBig),0,frac13sinthetaright) = \
&= left(Big(1+frac13costhetaBig)cosphi, Big(1+frac13costhetaBig)sinphi, frac13sintheta right)endalign

Define
$$h : S^1 times S^1 to mathbbR^3, h((x,y),(u,v)) = (x(1+fracu3),y(1+fracu3),fracv3) .$$

1) $h$ is injective.

Let $h((x,y),(u,v)) = h((x',y'),(u',v'))$. Then $v = v'$ and therefore $u^2 = 1 -v^2 = 1 - (v')^2 = (u')^2$. But now
$$(1+fracu3)^2 = (x^2+y^2)(1+fracu3)^2 = x^2(1+fracu3)^2 + y^2(1+fracu3)^2 = \ (x')^2(1+fracu'3)^2 + (y')^2(1+fracu'3)^2 = ((x')^2 + (y')^2)(1+fracu'3)^2 = (1+fracu'3)^2 .$$
This implies $u = u'$ and we conclude that also $x=x'$, $y=y'$.

2) $h(S^1 times S^1) = D$.

Let $C_(x,y)$ be the circle obtained by rotating $C_1$ around the $z$-axis until its center reaches $(x,y,0) in C_2$. It is contained in the plane spanned by $(0,0,0)$, $(x,y,0)$ and $(0,0,1)$. Obviously
$$C_(x,y) = a(x,y,0) + b(0,0,1) mid (frac13)^2 = lVert a(x,y,0) + b(0,0,1) - (x,y,0) rVert^2 \= lVert ((a-1)x,(a-1)y,b) rVert^2 = (a-1)^2 + b^2 .$$
Substituting $fracu3 = a-1$ and $fracv3 = b$, we get
$$C_(x,y) = (x(1 + fracu3),y(1 + fracu3),fracv3) mid (u,v) in S^1 = h((x,y) times S^1) .$$
1) and 2) show that $h$ gives the desired homeomorphism $S^1 times S^1 to D$.