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I'm trying to prove that $\pi_1(S^n)=0$ when $n>1$.

I've noticed that if you just take the north pole and south pole from each sphere, you can apply Van Kampen theorem. This doesn't fail as when $n>1$ it's still simply connected as it's homotopic to a circle.

But, how would you make this argument more formal? Would you use induction. But, can't see what the induction step will be. Help?

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You're almost there! You don't even need to use induction on dimension. You may use Van Kampen as you note ... so just write down what Van Kampen tells you. –  Neal Mar 11 '12 at 16:51
    
Notice also that you don't need the full force of Van Kampen's theorem: you only the easy part; that if $X$ is covered by open sets $U$ and $V$ with path connected intersection, the induced map $\pi(U)*\pi(V)\rightarrow \pi(X)$ is surjective. –  user17786 Mar 11 '12 at 21:27

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As you said you have to use Van Kampen theorem, but you don't need any induction argument. Let's consider a pair of antipodal point $P,Q \in S^n$ and consider the open sets $A=S^n \setminus \{P\}$ and $B=S^n \setminus \{Q\}$. The opensets $A$ and $B$ are both homemorphic to $D^n$ (the $n$-dimensional disk) and so are both simply connected, their intersection is path-connected, so by Van Kampen theorem $\pi_1(S^n)$ is quotient of $\pi_1(A)*\pi_1(B)=0*0=0$ (here $*$ is the free product of groups), thus $\pi_1(S^n)=0$.

Clearly this argument holds only for $n>1$ because just in this case the intersection of the open sets $A$ and $B$ is path-connected.

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Hey thanks yeah. I thought you had to use induction. But, yeah I suppose just need path connected to deduce that condition of Van Kampen theorem hold. Thanks. –  simplicity Mar 11 '12 at 16:57
    
You're welcome :) –  Giorgio Mossa Mar 12 '12 at 8:21

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