A sphere $S^{n}\subseteq \mathbb{R^{n+1}}$ is simply connected for $n\geq 2$ Show that the sphere $S^{n}\subseteq \mathbb{R^{n+1}}$ is simply connected when $n\geq 2$
Hints that I have are following :
$\textbf{Step 1}$
Every curve $\gamma:I \rightarrow S^{n}$ whose image is not the whole sphere (circle ) can be deformed continuously over the sphere (circle) to a single point by a simple geometric construction ! (Think of stereographic projection)
(Infact I have got that first step by considering the stereographic projection.)
So the only curves that make problems are (for $n\geq2$ ) the very pathological $\textbf{surjective}$ curves (so -called Peano curves)! 
$\textbf{Step2}$ Use Continuously deformation (homotopy) of continuous function into an arbitrary close $C^{\infty}$-function, applied to the component of a curve $\gamma :I \rightarrow S^{n}$ . The problem is to get a "homotopy" with values again in $S^{n}$. Reduce to step 1
Please help me ,May be its really basic but unfortunately I am not getting , I will be very grateful for it .  
 A: There are several ways to do this :


*

*You can find the null-homotopy explicitly. You can 'roll up' the closed curve to the north pole along great circles. (You may think this for $n=2$ case first.)

*Use Seifert Van Kampen theorem to compute $\pi_{1}(S^{n}) = 0$ by consider $S^{n}$ as union of two disks $D^{n}$ with intersection homotopic to $S^{n-1}$. 

*$S^{n}$ is homeomorphic to the one-point compactification of $\mathbb{R}^{n}$.
A: One of the ways to show that the sphere $S^n$ is simply connected when $n\geq2$ is to observe that it is isomorphic to the closed ball $D_n=\{x\in{\Bbb R}^n\,\mid\,|x|\leq1\}$ with the boundary identified to one single point, and the boundary is itself connected.
When $n=1$ the identification is still true, but in that case the boundary of the segment $[-1,1]$ is not connected.
A: We will show that if $k>n$, then $\pi_n(S^k)$ is trivial. This reduces to your question if we put $n=1$.
$S^i$ has an obvious CW-strucute given by one $0$-cell and one $i$-cell. Now if we are given any continuous map $f:S^n \to S^k$ where $n < k$, then by the cellular approximation theorem , $f$ is homotopic to a map whose image is contained in the $n$-skeleton of $S^k$. But the $n$-skeleton of $S^k$ consists only of that single $0$-cell, i.e. a single point. Thus $f$ is homotopic to a constant map and therefore $\pi_n(S^k)$ is the trivial group.
