Is there a value that appears exactly at one point?

Question

Let $S^2$ be the surface of the unit ball in $\mathbb{R}^3$ and let $T\colon S^2\rightarrow \mathbb{R}$ be a continuous map.

We have that $T$ gets a maximum $T_{\text{max}}$ and a minimum $T_{\text{min}}$ and that there is a point $x_0\in S^2$ with $T(x_0)=T(-x_0)$.

I want to check if there is value $T\in (T_{\text{min}}, T_{\text{max}})$ that appears at exactly one point on $S^2$.

So, for $h\neq 0$ we want to check if there is a point $x$ such that $f(x)=T(x)-T(x+h)$ has exactly one root, right?

To show that there is at least one root to we use the intermediate value theorem?

But how can we conclude that there is exactly one root? Could you give me a hint?

Answer 1

Definitely not. If $f:S^2\to\mathbb R$ is defined by $f(x_1,x_2,x_3)=x_3$, then $T_{max}=1=-T_{min}$, and for any $T\in (T_{\text{min}}, T_{\text{max}})$, we have $$f^{-1}(\{T\})=\{(x_1,x_2,T):x_1^2+x_2^2=1-T^2\}$$ is a cicle, hence has infinitely many points.

Now if you ask for $T\in [T_{\text{min}}, T_{\text{max}}]$, things change.