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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 homotopic to a constant map?

Thanks.

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    This is NOT a duplicate of the marked question. That question was asking whether such a mapping could be surjective, while this question asks why all such maps are homotopic to constant maps.2014-01-30

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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. If $h:|K|→|L|$ is a continuous function, then we know by the finite simplicial approximation theorem that there is a subdivision K' of $K$ and a simplicial map f:K'→L such that $h$ is homotopic to $f$. However, the image of $f$ is contained in the $n$-skeleton of $L$ (which is a proper subspace of $|L|$ since the dimension of $L$ is m>n). The proof is complete.

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.

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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.