Unit Spheres, Isotopies, and Homotopies

Question

I'm really struggling with the following problem.

Let $f:S^{2}\to\mathbf{R}^{3}$ be the embedding of the unit sphere, and let $E$ be the ellipsoid $E=\{(x,y,z)\in\mathbf{R}^{3}~:~4x^{2}+9y^{2}+z^{2}=1\}$. Parameterize $E$ by a map $g:S^{2}\to\mathbf{R}^{3}$, and show that $g$ is isotopic to $f$. Then prove that any map $f:S^{1}\to S^{n}$ is homotopic to a constant map $g:S^{1}\to S^{n}$.

For the first part, I'm not quite sure what the parameterization should be. I was thinking maybe $g(x,y,z)=(2x,3y,z)$, but I'm not sure that this is well-defined. Also, I know that an isotopy on a manifold $M$ is a smooth map $\Phi:[0,1]\times M\to M$ such that $\Phi_{t}:M\to M$ is a diffeomorphism for each $t\in[0,1]$. However, I'm not sure what my map $g$ should be.

For the second part, I know that a homotopy between $f:S^{1}\to S^{n}$ and $g:S^{1}\to S^{n}$ is a continuous map $H:S^{1}\times[0,1]\to S^{n}$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$ for all $x\in S^{1}$. I think the required homotopy might have something to do with stereographic projection, but that's about all I can come up with.

Any help is greatly appreciated!

Answer 1

For both of these, straight line homotopy is your friend! Do you see why stereographic projection will work in this context?

EDIT: The isotopy of $\mathbb{R}^3$ which takes $S^2$ to the ellipsoid is the straight line homotopy between the two. This ellipsoid is a compression of the sphere in two directions by different amounts - each original point can see where he's going with a straight line and none of these lines intersect. This is the idea behind the isotopy.

$$f_t(x,y,z)=t(1/2x,1/3y,z)+(1-t)(x,y,z)$$

At time $t=0$, $f_0$ is the identity map on $\mathbb{R}^3$. At time $t=1$, $f_1$ is the map which stretches in the $x$ and $y$ directions. This is essentially the solution.

For the second problem, to show that two maps are homotopic we have to build a map like above, except we aren't requiring that things can't pass through themselves. And, in fact, we're trying to show that something is homotopic to a constant map so we'll definitely have crushing somewhere. Stereographically project the image of $f:S^1\to S^n$ into $R^n$ by choosing a puncture point carefully. Use straight line homotopy like above, but connect everything to a single point. When you go backwards by the inverse projection map, this gives you your homotopy in $S^n$.

Answer 2

For the second part:

At first, suppose that there exists a point $x\in S^n$ such that $x\not\in f(S^1)$. Thus the image of $f$ is contained in $\left(S^n \setminus x\right)$, which is contractible. Therefore $f$ is null-homotopic.

In general, it is possible that $f(S^1) = S^n$ (see: Space-filling curve). You can show, however, that $f$ is homotopic to $\hat{f}$ such that $\hat{f}(S^1) \subset \left(S^n \setminus x\right)$ for some $x\in S^n$. Mainly, for some $x\in S^n$ and $\epsilon > 0$ take a ball $B(x,\epsilon) \subset S^n$ and for every piece of the curve $f(S^1)$ inside the ball replace it by an arc laying on the ball's boundary. You should be able to prove that $\hat{f}$ constructed this way is still homotopic to $f$. Since homotopy is a transitive relation, you can use the reasoning from the first case to prove that $f$ is homotopic to a constant map.