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Let $f:S^{n-1}\to \mathbb{R}$ be a nonconstant continuous function.

  1. $S^{n-1}$ is Compact & Connected$\implies$ Its image is of the form $[a,b]$ in $\mathbb{R}$.

  2. For any $c \in (a,b)$: $\#\{f^{-1}(c)\}>1$.
    If not, then $f:S^{n-1}\setminus \{f^{-1}(c)\}\to \mathbb{R}$ is continuous and since $S^{n-1}\setminus \{f^{-1}(c)\}$ is connected it image i.e $[a,c)\cup(c,b]$ is connected. Contradiction.

  3. There exist $x \in S^{n-1}$ such that $f(x)=f(-x)$
    Consider $g(x)=f(x)-f(-x) \implies g$ is an odd function. If $g(x)=0$ then we are done. If not, as $g(S^{n-1})$ is connected, either $g(x)>0$ or $g(x)<0$ for all $x \in S^{n-1}$. Contradicting the fact that $g$ is odd.

Geometrically I can see that:

  1. In $S^1$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ has atleast two points.

  2. In $S^2$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ should atleast contain a shape of homeomorphic to a circle i.e $S^1$ (Is this true?)

My Question: Can we tell that in $S^{n-1}(n\ge3)$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ contains a set homeomorphic to $S^{n-2}$?

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    $\begingroup$ Minor note: In point 1, you can't conclude that the image is of the form $[a, b]$ just from compactness. You are also using the fact that $S^1$ is connected. $\endgroup$ – Aryaman Maithani 2 days ago
  • $\begingroup$ @AryamanMaithani: Thank you for pointing it out. I have edited it. $\endgroup$ – Saikat 22 hours ago
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My Question: Can we tell that in $S^{n-1}(n\ge3)$ for every $c \in (a,b)$: $\{f^{-1}(c)\}$ contains a set homeomorphic to $S^{n-2}$?

No.

Consider any closed subset $A\subseteq S^{n-1}$ such that $A$ disconnects the sphere into two nonempty open components $U_1,U_2$. Since $S^{n-1}$ is perfectly normal then we can construct $f:S^{n-1}\to [-1,1]$ such that $f^{-1}(0)=A$ and $f^{-1}(-1)$ is a singleton in $U_1$ while $f^{-1}(1)$ is a singleton in $U_2$.

All we need to have now is a subset $A\subseteq S^{n-1}$ that is closed, disconnects the sphere but it does not contain a copy of $S^{n-2}$. For $n=3$ this can be simply done by taking the (closed) topologist's sine curve around $S^{2}$. This then can be lifted to higher dimensions.

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    $\begingroup$ Dang! This disconnecting curve is very close to being a loop; I wonder, can the conjecture be salvaged by requiring $f$ to be smooth? $\endgroup$ – Karl 2 days ago

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