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Is it true that any continuous map $\mathbb{S}^n\to \mathbb{S}^m$ is not surjective if $n<m$?

Thanks.

Aspirin
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  • they can be surjective, but are homotopic to nonsurjective maps (eg hatcher section 4.1) – yoyo Dec 26 '11 at 00:50

1 Answers1

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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 $n\geq1$.

Mariano Suárez-Álvarez
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    Aren't these discontinuous? – Alex Becker Dec 25 '11 at 23:37
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    Alex, the whole point of Peano curves is that they are continuous :) – Mariano Suárez-Álvarez Dec 25 '11 at 23:38
  • @Alex: No, they are continuous. – Jim Belk Dec 25 '11 at 23:39
  • They're curves, see also the The Hahn–Mazurkiewicz theorem, http://en.wikipedia.org/wiki/Peano_curves – Ehsan M. Kermani Dec 25 '11 at 23:41
  • @Mariano: and how we can proof that such map $\mathbb{S}^n\to\mathbb{S}^m$ is homotopic to a constant map? – Aspirin Dec 26 '11 at 00:01
  • @Aspirin: degree of such maps is zero, then use the Hopf theorem. – Ehsan M. Kermani Dec 26 '11 at 00:33
  • @Aspirin, you should ask that as a separate question. – Mariano Suárez-Álvarez Dec 26 '11 at 00:39
  • @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 $n<m$. If $h:\left|K\right|\to \left|L\right|$ 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. – Amitesh Datta Dec 26 '11 at 01:37
  • @Aspirin 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 $N$. – Amitesh Datta Dec 26 '11 at 01:42
  • @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. – Amitesh Datta Dec 26 '11 at 01:44