Is it true that any continuous map $\mathbb{S}^n\to \mathbb{S}^m$ is not surjective if $n Thanks.
Continuous map $\mathbb{S}^n\to \mathbb{S}^m$
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general-topology
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0they can be surjective, but are homotopic to nonsurjective maps (eg hatcher section 4.1) – 2011-12-26
1 Answers
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No, it is not true. There are variations of the Peano curve which provide surjective maps $S^1\to S^n$ for all $n\geq1$.
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1Aren't these discontinuous? – 2011-12-25
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1Alex, the whole point of Peano curves is that they are continuous :) – 2011-12-25
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0@Alex: No, they are continuous. – 2011-12-25
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0They're curves, see also the The Hahn–Mazurkiewicz theorem, http://en.wikipedia.org/wiki/Peano_curves – 2011-12-25
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0@Mariano: and how we can proof that such map $\mathbb{S}^n\to\mathbb{S}^m$ is homotopic to a constant map? – 2011-12-26
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0@Aspirin: degree of such maps is zero, then use the Hopf theorem. – 2011-12-26
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0@Aspirin, you should ask that as a separate question. – 2011-12-26
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0@Aspirin In summary, one proof is as follows: let us choose triangulations $h:\left|K\right|\to S^n$ and $k:\left|L\right|\to S^m$ where $K$ and $L$ are simplicial complexes and $n
$h:\left|K\right|\to \left|L\right|$ 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 $f:K'\to L$ 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 $\left|L\right|$ since the dimension of $L$ is $m>n$). The proof is complete. – 2011-12-26 -
0@Aspirin You might wish to look at the following Wikipedia articles: [Simplicial complex](http://en.wikipedia.org/wiki/Simplicial_complex), [Barycentric subdivision](http://en.wikipedia.org/wiki/Barycentric_subdivision), and [Simplicial approximation theorem](http://en.wikipedia.org/wiki/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 $N$. – 2011-12-26
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0@Aspirin 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. – 2011-12-26