$\pi_1(S^n)=0$ for $n\geq2$ Hello friends of math :D
I want to prove the result named in the heading. I have some hints but i can't imagine how wo work with this to conclude the result:


*

*Consider $S^n\subset\Bbb{R^{n+1}}$ (as metric space) and cover this by open balls of diameter $1$. Suppose $p:[0,1]\rightarrow S^n$ a path with $p(0)=p(1)$. Show that there exists a partition $0<t_0<t_1<\cdots<1$ such that $p([t_i,t_{i+1}])$ is contained in one of the balls for each $i$.

*Show that $p$ restricted to $[t_i,t_{i+1}]$ is homotopic to a path from $p(t_i)$ to $p(t_{i+1})$ along a great circle.

*Conclude that $p$ is homotopic to a path $p'$ consisting of finitely many arcs on great circles and conclude $p'[0,1]\neq S^n$

*Show that $p'$, thus $p$ is homotopic to a constant loop.


I have no idea. I find homotopy very difficult. Can someone help? Thanks :)
 A: I will help you out by proving the first hint and last hint. 


*

*Cover $S^n$ with balls of radius $1$. Now $S^n$ is a compact set so there exists a finite subcover $\{B_1(x_i)\}_{i=1}^n$ . Now for each $t \in [0,1]$ we have $p(t) \in B_i$ for some $i$ and choose an open neighbourhood $V_{p(t)}$ about $p(t)$ that is contained in $B_i$. Then continuity of $p$ tells us that there is $U_t$ open about $t$ such that $$p(U_t) \subseteq V_{p(t)} \subseteq B_i.$$
As $t$ runs over all points of $[0,1]$ the collection $\{U_t\}$ is an open cover of $[0,1]$ and thus by compactness we can extract a finite subcover $\{U_{t_i}\}_{i=1}^n$  where $t_i$ are points of $[0,1]$. Now there is no loss of generality in assuming a priori that your $U_t$ are open intervals. Hence there exists a partition
$$0 < t_0 < \ldots <t_{i} < 1$$
such that $p([t_i,t_{i+1}])$ is contained in one of the balls for each $i$.

*Suppose $p'$ is not the whole of $S^n$. Then there is $x \in S^n$ such that $x \notin \operatorname{im} p'$. Thus $p'$ is a loop in $S^n - \{x\}$ that is homeomorphic to $\Bbb{R}^n$. Since $\Bbb{R}^n$ is a convex set necessarily $p'$ is nullhomotopic.
A: *

*The partition exists because $[0,1]$ is compact

*The intersection of an open ball $B$ with the closed $n$-ball $D^n$ is convex, hence contains the straight line from $p(0)$ to $p(1)$, which does not pass through the origin. The projection of this line from the origin to the surface is (part of a) great circle and that is contained in the intersection  $B\cap S^n$. Moreover, in $D_n$ the given path from $p(0)$ to $p(1)$ is homotopic to the straight line. Projecting the homotopy gives a homotopy in $B\cap S^n$ as desired.

*clear

*$p'$ leaves out at least one point. Use stereographic projection to argument in $\mathbb R^n$.

A: You can generate the proof by this fact -which is easier to prove I think-:
Let $X$ be a topological space such that there exist two open and simply connected subsets $U$, $V$ $⊂ X$ such that the union is $X$ and $U∩V$ is path-connected, then X is simply connected.
to prove this just take an element in $U∩V$ and use Lebesgue lemma to find the partition in $[0,1]$ and join the images of the partition by the fixed element in the intersection ... you get the proof after some easy steps.
Now for $S^n$, $n>1$ choose two distinct points on the space $x,y$ and note that $S^n-{x}$ and $S^n-{y}$ are open cover for $S^n$ and they are simply connected because they are homeomorphic to $R^n$ and $S^n-{x}∩S^n-{y}$ is at least path-connected.. then $S^n, n>1$ is simply connected i.e $π_1(S^n)=0, n≥2$
