Possible Duplicate:
Continuous map $\mathbb{S}^n\to \mathbb{S}^m$
Why is every continuous function $f:\mathbb{S}^n\to\mathbb{S}^m,$ for $n
Thanks.
Possible Duplicate:
Continuous map $\mathbb{S}^n\to \mathbb{S}^m$
Why is every continuous function $f:\mathbb{S}^n\to\mathbb{S}^m,$ for $n
Thanks.
In summary, one proof is as follows: let us choose triangulations $h:\left|K\right|\to S^n$ and $k:|L|→S^m$ where $K$ and $L$ are simplicial complexes and $n
You might wish to look at the following Wikipedia articles: Simplicial complex, Barycentric subdivision, and Simplicial approximation theorem. In fact, the subdivision K' of $K$ in my proof above can be chosed to be the $N$th barycentric subdivision of $K$ for some nonnegative integer $N$.
A version of the simplicial approximation theorem remains true in the case where $K$ and $L$ are arbitrary (not necessarily finite) simplicial complexes. However, in the general case, one needs to consider subdivisions of $K$ more general than barycentric subdivision. The details underyling the ideas that I have presented here can be found in pages 79-99 of Elements of Algebraic Topology by James Munkres.
Unsurprisingly, I prefer a more analytic approach: since $f$ is uniformly continuous, for sufficiently small $\epsilon>0$ the mollification $f_\epsilon:=f*\phi_\epsilon$ satisfies $\sup|f_\epsilon-f|<1/5$. Let $g=f_\epsilon/|f_\epsilon|$ to return the values to the sphere. Since $\sup |f-g|<2/5$, there is a natural "straight-line" homotopy between $f$ and $g$, along the unique length-minimizing geodesic from $f(x)$ to $g(x)$.
Since $g$ is smooth, the $m$-dimensional measure of $f(\mathbb S^n)$ is zero, which implies $g$ is not surjective. Suppose it omits the North Pole: then we shrink it into the South pole along the longitude lines.