Using Mayer-Vietoris to find map whose degree is arbitrary natural number from $S^n \rightarrow S^n$

Question

Show that if $d$ is an integer and if $n \geqslant 1$, then there is a map $h: S^n \rightarrow S^n$ of degree $d$.

So I'm starting off by writing $S^n$ as the union of it's top and bottom hemispheres, $T$ and $B$ so that $T \bigcap B = S^{n-1}$. Then I'm looking at the induced homology sequence whilst using an induction hypothesis on $n$, but from there i'm a bit lost.

Any insight is appreciated!

Answer 1

Let $S^n=S^{n-1}\times[-1,1]/\sim$ where the equivalence relation identifies all the points in $S^{n-1}\times\{-1\}$ and $S^{n-1}\times\{1\}$. Write $S^n$ as the union $U=S^{n-1}\times(-1,1]$ and $V=S^{n-1}\times [-1,1)$. Note that $U$ and $V$ are contractible and their intersection is homotopy equivalent to $S^{n-1}$. Suppose we have a map $f_{n-1}:S^{n-1}\rightarrow S^{n-1}$ of a certain degree d. Then $f(x,t)=(f_{n-1}(x),t)$ preserves $U$ and $V$ and their intersection. The Mayer Vietoris sequence is natural so we will get induced maps betwee the sequences. This boils down to the following diagram:

$\require{AMScd}$ \begin{CD} H_n(S^{n}) @>>> H_{n-1}(U\cap V)\\ @V V f_n V @VV f_{n} V\\ H_n(S^{n}) @>>> H_{n-1}(U\cap V) \end{CD}

Everything in sight commutes and the horizontal maps are isomorphisms. On the right this is nothing but the degree $d$ map $f_{n-1}$ up to homotopy equivalence. The conclusion is that $f_n$ must also have degree $d$.

So the question is if you can produce a degree d map for $n=1$. I'm sure you can find many references.

Answer 2

Sometimes it helps to just write things down.

For $x = (x_1, \ldots, x_{n+1}) \in S^n - K$, where $K$ is the set of points of the form $(0, 0, t_3, \ldots t_{n+1})$, define $$ \theta(x) = atan2(x_2, x_1)\\ r(x) = \sqrt{x_1^2 + x_2^2}. $$

(roughly: longitude, and radius of the "sphere of latitude" on which $x$ lies)

Now define $$ f_k(x) = \begin{cases} ( r(x) \cos (k \cdot \theta(x)), r(x) \sin (k \cdot \theta(x)), x_3, \ldots, x_{n+1}) & x \notin K \\ x & x \in K \end{cases} $$

Then $f_k$ is a degree-$k$ map, as you can see by counting the preimage of a typical point (i.e., any point not in the measure-zero set $K$).

[By the way, this is simply the "suspension" map that others have mentioned, but written out explicitly so that it's really easy to work with.]