$S := \{x \in \Bbb R^3: ||x||_2 = 1 \}$ and $T: S^2 \to \Bbb R$ is a continuous function. Is $T$ injective? $S := \{x \in \Bbb R^3: ||x||_2 = 1 \} \subset (\Bbb R^3, || \cdot ||_2 )$ and $T:S^2 \to (\Bbb R, |\cdot |)$ is  a continuous function.
I've already shown that $$T_{\mathrm{max}} := \mathrm{sup}\{ T(x) : x \in S^2\}  \: \: \mathrm{ and }\: \: T_{\mathrm{min}}:= \mathrm{inf}\{ T(x) : x \in S^2\}$$ are reached by $T$ with the argument that $S^2$ is closed and bounded and therefore after the Heine-Borel-Theorem compact, and therefore since T is continuous it assumes its max and min in $S^2$. 
Now i have answer the following:

1) Is there a point $x_0 \in S^2$ with $T(x_0)=T(-x_0)$ ?
2) Is there a value $T \in(T_{\mathrm{min}}, T_{\mathrm{max}})$ which is only assumed at exactly one point of $S^2$ ?

I wanted to show that $T$ is injective to get that 1) is false and 2) is true but i didn't get anywhere.
Any ideas? Thanks in advance!
 A: $T$ cannot be injective. If $T$ is constant this is obvious, so assume $T$ is not constant. Then, as you explained already, $T$ attains its minimum an maximum in $x_0$ and $x_1$, say, respectively. Now choose two different paths from $x_0$ to $x_1$ which are disjoint with the exception of the common endpoints. The image of each covers the interval $[T(x_0), T(x_1)]$ by elementary (one dimensional) arguments.
Edit (to answer a comment and give a complete answer to the question):
The claim ('one dimensional argument') follows parametrizing the curves, so you get  continuous maps $[0,1]\rightarrow \mathbb{R}$ which allows to apply the intermediate value theorem. With this reasoning it also follows that the answer to 2. is 'no'
to answer 1.), consider $f(x):= T(x)-T(-x)$ and note that $f(-x) = T(-x) - T(x) = -f(x)$. So either $f$ is identically $0$ which answers the question trivially, or, if $f$ is negative in some point $a$ it will be positive in the antipodal point $-a$. Now consider any curve $c:[0,1]\rightarrow S^2$ joining $a$ and $-a$ and look at $f\circ c$. By the intermediate value theorem, this will attain the value $0$, which implies there is a point $b$ such that $T(b)=T(-b)$
