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Serre proved, that for (allmost) all $n,m\in\mathbb{N}$ the homotopy groups $\pi_n(\mathbb{S}^m)$ are finite, so - using simplicial approximation - for $n, m$ fixed there is a finite cell decomposition $\Sigma^n$ of $\mathbb{S}^n$ such that all maps $f:\mathbb{S}^n\rightarrow\mathbb{S}^m$ can be represented (up to homotopy) by a simplicial map $\mathfrak{f}:\Sigma^n\rightarrow \mathfrak{X}^m$ for any given (probably easy) simplicial structure $\mathfrak{X}^m$ on $\mathbb{S}^n$.

Are there any known bounds on the size of such a simplicial decomposition and can these be used to gain information about the $\pi_n(\mathbb{S}^m)$?

Note, that this is a duplicate of https://math.stackexchange.com/questions/2884130/how-many-cells-do-we-need-in-mathbbsn-to-induce-pi-n-mathbbsm, where I was adviced to post this question here.

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    $\begingroup$ Your question sits in a family of similar questions. I recall someone (?perhaps Mike Freedman?) asked a similar question but where he used some kind of elasticity measure of the smooth map, rather than simplicial structures. i.e. you're trying to make some kind of measurement of how complicated the map is, at some level. I can't find that old MO thread. $\endgroup$
    – Ryan Budney
    Commented Aug 15, 2018 at 22:10
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    $\begingroup$ This question is also related (but not quite the same): mathoverflow.net/questions/285632/… $\endgroup$
    – j.c.
    Commented Aug 15, 2018 at 22:58
  • $\begingroup$ I am interested in a related question also when n=m. In this case the homotopy group is the integers (which is infinite), and so we cannot expect a single cell decomposition on the domain to work for all maps f, which are determined (up to homotopy) by their degree. Nevertheless, we can fix a degree d, and ask what is a "minimal" simplicial decomposition of the domain admitting a map of degree d. $\endgroup$
    – Henry Adams
    Commented Sep 5, 2018 at 12:14

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