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.

Take an injective (continuous) map $T:\mathbb{S}_1\to\mathbb{S}_1$. It's an obvious fact (though I can only prove it with nontrivial facts about homotopy) that $T$ is automatically surjective.

I have two questions:

  • Is this the case for all 'endomorphisms' of spheres? (I think so, but have not proven it)
  • Does there exist a compact manifold without boundary which embeds nontrivially into itself?

I suspect the answer to the second question is 'no', but I have little evidence for the conjecture.

share|improve this question
6  
$\newcommand{\ra}{\rightarrow}$If $f: S^n \ra S^n$ misses a point then by stereographic projection it induces a map $f: S^n \ra R^n$ so if there were an injective non-surjective map from $S^n$ to $S^n$, then $S^n$ would embed in $R^n$. For your second question you might try creating a descending chain by repeated applications of your embedding or something along those lines. –  JSchlather Oct 11 '12 at 3:37
    
Thinking about the degree of such a map should be profitable. If a map misses a point, it must have degree zero. I'm not sure if this is enough to rule out an embedding, though. –  NKS Oct 11 '12 at 3:57
1  
The second question is interesting because both "compact" and "without boundary" are essential, and it's hard to see how they interact. –  Nate Eldredge Oct 11 '12 at 4:02
7  
The second question follows (negatively) from the theorem of invariance of domain (en.m.wikipedia.org/wiki/Invariance_of_domain#section_3), which says that that such a map is open (and since your manifold is compact, the image is closed and open, and solves the problem for connected manifolds, then you use that there are finitely many connected components) –  user17786 Oct 11 '12 at 7:25
    
Thanks for the reference - very cool :) –  user33750 Oct 16 '12 at 0:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.