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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.

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    $\begingroup$ $\newcommand{\ra}{\rightarrow}$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. $\endgroup$
    – JSchlather
    Oct 11 '12 at 3:37
  • $\begingroup$ 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. $\endgroup$
    – NKS
    Oct 11 '12 at 3:57
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    $\begingroup$ The second question is interesting because both "compact" and "without boundary" are essential, and it's hard to see how they interact. $\endgroup$ Oct 11 '12 at 4:02
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    $\begingroup$ 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) $\endgroup$
    – user17786
    Oct 11 '12 at 7:25
  • $\begingroup$ Thanks for the reference - very cool :) $\endgroup$
    – user33750
    Oct 16 '12 at 0:35
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This is a CW answer to record that the question is answered in the comments. If someone upvotes this, the question will be removed from the unanswered list (and I will not receive reputation, because it is CW).

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