Q:

Continuous map on $S^2$

Can you help me with this?

Let $S^2 := \{x\in \mathbb R^3:||x||_2 = 1\} \subset (\mathbb R^3, ||\cdot||_2)$  and $T:S^2 \to (\mathbb R, |\cdot|)$ be a continuous map.
  a) Why does T assume its maximum ($T_{max}$) and its minimum ($T_{min}$)?
  b) Is there a point $x_0 \in S^2$ with $T(x_0) = T(-x_0)$?
  c) Is there a value $T\in]T_{min},T_{max}[$ that is assumed at only one $x\in S^2$?  

a) is easy, it is compact since it's closed and bounded (Heine-Borel).
For b) I know it can't be injective since it maps from a higher to a lower dimension, but I don't know if I can use that to show what I need.
For c) I assume I can use non-injectivity again, but don't know either how. Any hints?

A:

You can show b) using contradiction.
Assume that for all $x \in S^2$, $T(x) \neq T(-x)$.
Let $x_{max} \in S^2$ such that $T(x_{max})=T_{max}$
and $x_{min} \in S^2$ such that $T(x_{min})=T_{min}$
Let $f(x) = T(x)-T(-x)$.
By assumption, $f(x_{max}) >0$ and $f(x_{min}) <0$ 
$f$ is continuous on $S^2$ which is connected so by the intermediate value theorem, there exists $x_0 \in S^2$ such that $f(x_0)=0 \Leftrightarrow T(x_0)=T(-x_0)$. Contradiction.

