3

Let $f: S^2 \to \mathbb{R}^n$ be a smooth map from the two-dimensional sphere to euclidean space. Let $X = \mathrm{Im}(f) \subset \mathbb{R}^n$ be the image topological space (note: the quotient topology induced from $S^2$ and the subspace topology induced from $\mathbb{R}^n$ coincide here). I am interested in knowing what properties this topological space has.

Ideally, I would like $X$ to admit the structure of a $2$-dimensional CW-complex. From the comments in this question, this seems highly unlikely. But maybe it is at least homotopic to a CW-complex. To this end, I came across the notion of Euclidean neighbourhood retracts. The characterisation at the end of these slides imply that it suffices to prove that $X$ has sufficient local connectivity properties. But I have no idea how to prove this.

I'm not quite sure where to look for answers to these questions and so any references to relevant literature would be welcome.

unknownymous
  • 511
  • This is more appropriate for math.stackexchange.com. The answer to your "homotopic" question is negative, see https://math.stackexchange.com/questions/523416/is-there-any-example-of-space-not-having-the-homotopy-type-of-a-cw-complex (the Hawaiian earrings I use can be realized as the image of a smooth map of an interval, hence, of the sphere). – Moishe Kohan Mar 30 '24 at 01:14
  • @MoisheKohan Oh thanks. Thats interesting. What is the map from the interval $[0,1]$ to the Hawaiian earrings? – unknownymous Mar 30 '24 at 01:28
  • The image won't have the homotopy-type of a CW-complex unless you put some regularity conditions on the map. For example, if you demand the map is an immersion, or you could allow the derivative to degenerate but only in the Whitney-style singularities, etc. The problem is that smooth maps are too close a class to continuous maps. – Ryan Budney Mar 30 '24 at 01:32
  • To get your map to the Hawaiian earrings, just "follow your nose". Divide the interval $[0,1]$ into a union $[0,1/2]$, $[1/2,3/4]$, $[3/4,7/8]$, etc and have each successive interval be the characteristic map of successive and smaller rings in the Hawaiian earrings. – Ryan Budney Mar 30 '24 at 01:39
  • Another useful exercise as you explore the possibilities of smooth maps is to construct a $C^\infty$-smooth map $\mathbb R \to ([0,\infty) \times {0}) \cup ({0} \times [0,\infty))$ that is one-to-one and onto. – Ryan Budney Mar 30 '24 at 01:42
  • Setting the "homotopic" question aside, what would be a compact subset of $\mathbb R^n$ that isn't the image of a smooth map $S^2\to\mathbb R^n$? – Igor Belegradek Mar 30 '24 at 01:44
  • @IgorBelegradek a two-point space. – Ryan Budney Mar 30 '24 at 01:45
  • 1
    @RyanBudney: okay, but can one say about such an image beyond trivialities? I think it is a reasonable question for MO. – Igor Belegradek Mar 30 '24 at 01:47
  • 1
    @IgorBelegradek: pathwise connected with Lebesgue covering dimension at most $2$, I would imagine that's the main constraint. – Ryan Budney Mar 30 '24 at 01:51
  • 4
    @IgorBelegradek: Any Peano continuum of Hausdorff dimension $>2$ is not the image of a Lipschitz map of $S^2$. – Moishe Kohan Mar 30 '24 at 01:59
  • I had the impression these kinds of questions were all answered well before the 60's, but I have not looked at the "dimension theory" literature in some time. – Ryan Budney Mar 30 '24 at 02:02

0 Answers0