Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n<m$ homotopic to a constant map?

Thanks.

share|improve this question

marked as duplicate by Amitesh Datta, Asaf Karagila, userNaN, J. M., t.b. Aug 15 '12 at 8:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
That is not true. For instance, there are space filling curves $S^1\rightarrow S^2$. What is true is that any map $S^n\rightarrow S^m$ is homotopic to a map satisfying that condition. –  Joe Johnson 126 Jan 14 '12 at 14:02
    
@Aspirin Could you please look at my comments here which were, in fact, directed to you? For your convenience, I have reproduced them as an answer below. –  Amitesh Datta Jan 14 '12 at 14:16
    
@Amitesh Datta: I'm sorry, I forget for this thread –  Aspirin Jan 14 '12 at 14:30
    
Dear @Asprin: I just thought I would point that out; it is not a problem. I think that you can either change the question to "Why is every continuous function $f:S^n\to S^m$ for $n<m$ homotopic to a constant map?" or flag the question for moderator attention (because, at the moment, this question is an exact duplicate and I think should be closed). –  Amitesh Datta Jan 14 '12 at 14:33
    
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. –  Donkey_2009 Jan 30 at 8:15

2 Answers 2

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 $n<m$. 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.

share|improve this answer

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.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.