Exercise 17, Chapter 7

Vasco,

I think when I wrote "This particular application of $H$ to two points $p$ and $p*$ is an instance of the mapping of $f$ to $p$ and $-p$ on the circle in exercise 16, because in this case, $p* = -p$", I forgot to preface it with something like "We can rotate the sphere so that any two opposite points lie on $\mathbb{C}$, where we can regard them as complex numbers." Right now, I don't see anything obviously wrong with it, but, I'm having misgivings, so I will spend some more time with it.  I realize that $H$ is spherical to complex, but I assumed that since stereographic projection is conformal, it also has an inverse. Otherwise, what do you do with "The theorem amounts to showing that $F$ has a root somewhere on S"?

You could be right to question the hint, since it seems odd to say that a function from $S$ to $\mathbb{C}$ would have a winding number, but in my solution it is also a function from $\mathbb{C}$ to itself, so it does wind on $\mathbb{C}$. And it seems no more strange than a root on $S$!

I have added a second, different, answer.

Gary

Vasco,

I was only arguing that $H(-p) = -H(p)$ in the special case where $p$ and $-p$ are on the equator. In that case I think, whether the projection is stereographic or vertical, $H(p) = p$.

Gary

Vasco,

I think the case is on the equator because the hint instructs us to take the equator as the boundary of the circle $C$ in exercise 16. It also seems like a valid method to rotate the antipodes to the equator to simplify the problem. It's a rigid movement. It would also be difficult to consider just the northern hemisphere unless the antipodal points were on the equator. This would make $H(p)$ a fixed point, a fact from which I have not drawn any further conclusions.

The metaphor of the deflating balloon suggests a vertical projection so I would not give the stereographic projection further consideration, but stereographic and vertical projections satisfy the requirement that H be a continuous mapping from $S$ into the plane.

Why do you think that $H(-p) = -H(p)$ is invalid? Given that H is merely continuous, I can see that it is not generally valid, but with vertical projection and antipodes on the equator, it seems valid.

I've made small changes to my answer.

Gary

Vasco,

$H$ is described as the deflation of $S$. On further consideration, I agree that we can't assume it is a vertical projection. We know that $S$ shrinks gradually, and I think we can assume uniformly, so we could deduce that points on a meridian will go to points on a ray, points on lines of latitude will go to points on concentric circles around the origin, angles between meridians and lines of latitude are preserved as angles between rays and concentric circles around the origin, and perhaps most significantly and with the most confidance, points on the equator are fixed. If antipodes are rotated to the equator prior to shrinking, they will be fixed points such that H(-p) = -H(p), so we can apply the results of ex. 16, that $\nu[H(p), 0] = odd$.

Gary

Vasco,

I agree. Your solution looks like what Needham must have had in mind.

Gary