I've asked this question about two month ago in math exchange but there were no answer to it. Any information or paper relating to this question is appreciated.

By the Borsuk-Ulam theorem we know that every continuous map $M:S^n\to \mathbb{R}^n$ has a point in its range that is the image of two points in the domain. By considering this, I am curious to see what amount of this double points would exist for such maps and what we can say about the set of points in the range which are the image of one or odd number of points in the domain for all maps.

It is known that for every continuous non-one element range map $f_1:S^1\to \mathbb{R}$ there are properties which state that for an even $e\in \mathbb{N}$ the set of points in its range which their preimage have exactly $e$ points is uncountable and for evey odd $o\in \mathbb{N}$ the $o$-point set in the codomain $\mathbb{R}$ has countable elements.

Also intuition maybe says something similar but more complex about two dimensional continuous maps $f_2:S^2\to \mathbb{R}^2$, where the dimension of $f_2(S^2)$ is $2$ (the same as codomain $\mathbb{R}^2$ dimension). It states that the $o$-point sets in $\mathbb{R}^2$ where $o\in \mathbb{N}$ is odd have one dimensional loop structure and for at least one even $e\in \mathbb{N}$ the $e$-point set has a two dimensional disc structure for each of its connected components.


First: can anyone present more rigorous statements about the latter case above? (with proof)

Second: In general does there any invariant of the continuous maps relate to aggregation of characteristics info (structure, size, dimension ,...) of $n$-point ($\forall n\in \mathbb{N}$) and uncount-point sets of the map?

Note: we define $n$-point set for a map $f:X\to Y$ is the set of all points such $p\in Y$ that the preimage $f^{-1}(\{p\})$ consists of exactly $n$ points in $X$, if the preimage $f^{-1}(\{p\})$ consist of uncountable points in the domain then we call it a uncount-point.

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    $\begingroup$ For a typical continuous map $f\colon\mathbb{S}^1\to \mathbb{R}$, the preimage of $f^{-1}\{x\}$ typical point $x$ is either empty or a Cantor set. Doesn't it contradict your statement? (I was not able to see its precise meaning) $\endgroup$ – Anton Petrunin Feb 25 at 5:47
  • $\begingroup$ @AntonPetrunin sorry I can't understand of what you call "typical". $\endgroup$ – MasM Feb 25 at 19:50
  • $\begingroup$ @AntonPetrunin I see it now ,good question, I think what you mentioned may add additional info about uncount-point set for a typical continuous function that should be considered from the point of question 2. $\endgroup$ – MasM Feb 25 at 21:39

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