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Write the $n$-sphere as the set $S^n\approx \{(x,z)\in\Bbb R^{n-1}\times\Bbb C: |x|^2+|z|^2=1\}$, and define a mapping $f: S^n\to S^n$ by $f((x,z)) = (x,\frac{z^2}{|z|})$. Why is $\deg(f)=2$?

Background: This is the essential content of a problem on one of my recent exams. I feel I am very close to solving this problem, but there is something missing. Here is the work so far:

Note that $p=(0,0,\dots, 0,1)$ is the image of exactly two points in $S^n$: these are $x_1=(0,0,\dots, 0,1)$ and $x_2=(0,0,\dots, 0,-1)$, and moreover it is contained in a neighborhood, each point of which is the image of exactly two points: for instance the inverse image of $S^n\cap \{(x,z): |x|<\varepsilon, \text{Re}(z)>0\}$ has inverse image $U_1\cup U_2$:

  • $U_1=S^n\cap \{(x,re^{i\theta}): |x|<\varepsilon \text{ and } -\frac{\pi}{4}<\theta<\frac{\pi}{4}\}$
  • $U_2=S^n\cap \{(x,re^{i\theta}): |x|<\varepsilon \text{ and } \frac{3\pi}{4}<\theta<\frac{5\pi}{4}\}$

We may therefore apply the local degree formula, and thus we need to know the action of $f$ that is induced on $H_n(U_i,U_i\smallsetminus x_i)\to H_n(V,V\smallsetminus p)$ for $i=1,2$.

Since $f$ is a homeomorphism $U_i\to V$, I know the local degrees are either $\pm 1$. But I don't know how to exclude the $-1$ case.

[Note: I'm pretty sure that $f$ can be described by repeatedly suspending $S^1\to S^1$. But that argument has some technicalities which I don't really understand, so I am trying to avoid it.]

Eric Stucky
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    To make sense of degree, you need to orient $S^n$, and check that this map is orientation preserving at both points. Your description of $f$ as $\Sigma^{n-1} s$ where $s: S^1 \to S^1$ is the squaring map is on point, and might be helpful for this, even if you don't want to use results that say that the degree of a suspension is the degree of the original map. Indeed, you'd be writing down a correct proof, no doubt. –  May 29 '16 at 01:10
  • @Mike: I think my issue is that I don't know how to define an orientation properly; I need to choose one of the two generators for $H_n(U_i,U_i\smallsetminus x_i)$. But what do these generators look like? Any time I try to start with chains, I can't think straight; is there a smarter way? Presumably they're not arbitrary; they're induced from the generator of $S^n$; but I don't know how this happens either. – Eric Stucky May 29 '16 at 01:33

1 Answers1

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You can think $S^n $ as iterated suspension of $S^1$. And with this identification the map will be the iterated suspension map of $S^1\rightarrow S^1 $sending $z\rightarrow z^2$.Hence degree is 2.

Gora
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