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How would you show that $\pi_n, n>1$ of the Klein bottle is the trivial group?

I was thinking Seifert-Van Kampen could be applicable?

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Julian Kuelshammer Nov 15 '12 at 17:29
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The universal cover of the Klein bottle is $\mathbb{R}^2$. You can use lifting criteria from covering space theory to show that the Klein bottle and its covering spaces have the same higher homotopy groups.

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So, then since $\pi_n(\mathbb{R}^2)=0$ for $n>1$, the result would follow, right? –  gabriel Nov 15 '12 at 17:37
    
@gabriel That's right. –  Joe Johnson 126 Nov 15 '12 at 18:26
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