Show that $T$ contains at most two points Take $n \geq 2$ and $f:S^{n} \to \mathbb{R}$ a continuous map.Call $T$ the set of points $t\in f(S^{n})$ such that the fibre $f^{-1}(t)$ has finite cardinality.
Show that $T$ contains at most two points.
I try understand the problem, but I can´t make a example of the proposition I try functions like projections and $x+y+z$, but I don´t get catch the statement of the proposition.
Any example that helps to me to see the proposition or in other form help me to catch the idea behind of the proof
 A: HINT: $S^n$ is compact and connected, so $f[S^n]$ must be a closed interval $[a,b]$. If $|T|\ge 3$, there must be some $t\in T\cap(a,b)$. Consider the sets $f^{-1}\big[[a,c)\big]$ and $f^{-1}\big[(c,b]\big]$.
A: Here's an example: view $S^2 \subset \mathbb{R}^{3}$ in the standard way, and consider the height map $(x, y, z) \mapsto z$. The fibers are parallels that are the same height above the $xy$-plane. The fibers $f^{-1}(1)$ and $f^{-1}(-1)$ each consist of one point, namely the poles $\{(0, 0, 1)\}$ and $\{(0, 0, -1)\}$, whereas every other fiber has an infinite number of points (such as $f^{-1}(0)$, which is the equator or unit circle in the $xy$-plane).
So in this case, there are exactly two values in $\mathbb R$ with finite cardinality preimages.
A: Let's first take one of you're attempted examples and work through it to see what's happening: take the 2-sphere $S^2 \subseteq \mathbb{R}$ and consider the projection onto the $z$-co-ordinate. The image of this map is $[-1,1]$.
Now at $-1$ and $1$, the only points that are sent to each of those by this projection map are the poles: $(0,0,-1)$ and $(0,0,1)$ respectively. So these two points both have a finite preimage.
But for any $z$ in between $-1$ and $1$, looking at the preimage of these from the projection map means taking an entire slice through the sphere at that height. You get an entire circle, all at the same height, so all sent to the same $z$ when you project out. All of these $z$s have an infinite set of points sent to them.
Thus for this projection map, exactly two points have finite preimage: $-1$ and $1$.

How are we going to prove that this works for any sphere with $n \geq 2$, and any continuous map?
Start off by noting that the sphere is closed and bounded subset (thus compact) and also connected. Therefore the image has to be a closed bounded interval $[a,b]$.
Can you show that $a$ and $b$ are the only two numbers that can have finite preimage? That is, knowing that there are points on the sphere sent to $a$ and $b$, can you show that for all the numbers between $a$ and $b$ there are infinitely many points on the sphere that are sent to that number?
A: Well the image of $f$ is going to be a connected compact subset of $\mathbb{R}$, so it will be a closed interval $[a,b]$ (or 1 point if $f$ is constant).
So it would be natural to think that $a$ and $b$ are the only possible values of $t\in[a,b]$ such that $f^{-1}(t)$ is finite. To prove this, let $p,q$ be points of $\mathbf{S}^n$ with $f(p)=a,f(q)=b$.
There are clearly infinite different paths $g_k:[0,1]\to\mathbf{S}^n$ in the sphere that go from $p$ to $q$. We can pick them so that they only meet at $p$ and $q$ and don´t share any other points (take for example circunference arcs that go from $p$ to $q$ in $\mathbf{S}^n$). Moreover, by the intermediate value theorem, $f\circ g_k:[0,1]\to[a,b]$ has $f\circ g_k(0)=a$ and $f\circ g_k(1)=b$, so $f\circ g_k$ must take the value $t$ at some point if $t\in(a,b)$ (thus, $f$ takes the value $t$ at some point of the path). That is, for given $t\in(a,b)$, you have at least one point $x$ in each path from $p$ to $q$ such that $f(x)=t$, so there must be infinitely different such points, showing that $f^{-1}(t)$ is infinite.
