2
$\begingroup$

Let $S^n$ be the sphere of dimension $n$. In order to construct a map $f:S^n\rightarrow S^n$ of degree $d\geq 2$ one has the following construction: Let $K$ be the complement of $d$ disjoint n-dimensional open discs in $S^n$. Then collapsing $K$ to a point we find that $S^n/K\simeq \bigvee _{i=1}^d S^n$. Identifying (choosing a homemomorphism) each of the sphere in the disjoint union with $S^n$ (with the approriate orientation), the composition of the two maps $$ S^n\rightarrow \bigvee _{i=1}^d S^n \rightarrow S^n, $$ gives us a map $\phi_d:S^{n}\rightarrow S^n$ of degree $d$.

In general, if $f:S^n\rightarrow S^n$ is a map of degree $d$ and $x\in S^{n}$ is such that the fiber $f^{-1}(x)$ is finite, then one has from excision theorem that $$ \sum_{y\in f^{-1}(x)} deg_f(y)=d. $$

Q1: How would you construct a (continuous) map $f:S^n\rightarrow S^n$ of degree $d$ such that on a dense subset of points $X\subseteq S^{n}$ one has that for every $x\in X$ that the fiber $f^{-1}(x)$ is infinite?

Q2: Having a map $f$ as in Q1 and a point $x\in X$, is it possible to take some kind natural average sum over the local degrees of the elements of $f^{-1}(x)$ in such a way that the sum converges to $d$ (you may assume that $S^n$ is endowed with a metric if you think it helps) ?

added: Note that if one constructs a map $f:S^n\rightarrow S^n$ of degree one with infinite fibers (on a dense set) then by post-composing with a map of degree $d$ we obtain a map of degree $d$ which satisfies all the conditions.

Improve this question
$\endgroup$
4
  • 5
    $\begingroup$ Think about the following example $f\colon \mathbb S^1\to \mathbb S^1$ defined as $$f(x)=x+\sum_n \tfrac1n\sin (n^2\cdot x) \pmod{2\cdot\pi}.$$ Note that the preimage $f^{-1}(y)$ for any $y$ is a Cantor set. $\endgroup$
    – Anton Petrunin
    Commented Oct 28, 2012 at 19:37
  • 1
    $\begingroup$ In the actual Q1 (in contrast to the title), you only asked that some point $x$ have an infinite fiber. The example you gave at the beginning of the question already does that. The set $K$ that you collapse to a point is one of the fibers of your map. $\endgroup$
    – Andreas Blass
    Commented Oct 29, 2012 at 0:35
  • $\begingroup$ Yes you are right, I had in mind something different. $\endgroup$
    – Hugo Chapdelaine
    Commented Oct 29, 2012 at 14:57
  • $\begingroup$ @Anton, I have 2 simple questions: what is the degree of your map $f$ (and how to compute it) and how do you compute a fiber $f^{-1}(y)$? $\endgroup$
    – Hugo Chapdelaine
    Commented Oct 29, 2012 at 15:11

1 Answer 1

Reset to default
1
$\begingroup$

Q1: It seems to me that you could take a disc $B_2$ in $S^2$ and a map $f: S^2\to S^2$ of a given degree s.t. $f(B_2)$ is a point $y$, and define $g(x)=f(x)$ for $x\notin B_2$ and $g(x)|_{B_2}$ to be the composition $B_2\to B_2/S^1\simeq S^2\to S^3 \to S^2$ of a factor map, an onto map $S^2\to S^3$ and the Hopf fibration such that $g(S_1)=y$. Each point in $S^2$ has infinite fiber and $g$ is homotopic to $f$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .