# Showing higher homotopy groups of $S^1$ are trivial

I'm trying to prove $\pi_{i} (S^1) \cong 0$ if $i>1$.

Is this correct.

You have a short exact sequence, $\mathbb{Z} \rightarrow \mathbb{R} \rightarrow S^1$ (from the fiber bundle of the covering space), then you deduce from some magic I don't understand you can do this $\pi_{i}(\mathbb{Z}) \rightarrow \pi_{i}(\mathbb{R}) \rightarrow \pi_{i}(\mathbb{R}) \rightarrow\pi_{i-1}(\mathbb{Z}) \rightarrow \ldots$

Is this correct?

I was wondering about $\pi_{i}(\mathbb{R})$ being trivial when $i>1$. But, yeah made it unclear as I was asking two questions.

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$\mathbb R$ is contractible. – Rasmus May 15 '12 at 16:51
But, I only know the proof that involves $\pi_{1}(X)$. Don't know how to show the above top. Don't know if something destroys the contraction if you take higher homotopy. – simplicity May 15 '12 at 16:59
Are you really trying to prove $\pi_i$ is a functor? If by $X \simeq Y$ you mean that $X$ is homeomorphic to $Y$, isn't it easy to just prove directly that $\pi_i(X)\simeq\pi_i(Y)$ using the definition if $\pi_i$ and using composition with $\phi$ and $\phi^{-1}$ where $\phi$ is a homeomorphism from $X$ to $Y$? – Thomas Andrews May 15 '12 at 17:30
You write "So I'm trying to prove that $\pi_i(X)\simeq\pi_i(Y)$ if $X\simeq Y$. I want to show that $\pi_i(S^1)\simeq 0$ for $i>1$." Which do you want to show? These two are entirely different problems - one result does not in any way naturally lead to the other. (And neither of these results reflect your original title.) – Thomas Andrews May 15 '12 at 18:29
@SteveD: I agree the question is unclear, but the exact sequence looks correct to me. (It's a fragment of the long exact sequence of a fibration.) – Grumpy Parsnip May 15 '12 at 19:11

This is not the standard way to prove the higher homotopy groups of $S^1$ are trivial, but I'll bite.

First, a series of lemmas.

Lemma 0: Each connected component of $\mathbb{Z}$ (i.e, each point) is contractible - the identity map, restricted to each component, is homotopic to a constant map.

Proof: The connected components are points, so the identity map is constant.

Lemma 1: $\mathbb{R}^n$ is contractible, that is, the identity map is homotopic to a constant map.

Proof: Consider the homotopy $F(x,t) = tx$. When $t=0$, this is a constant map and when $t=1$, it's the identity map. (I'll leave it to you to prove $F$ is continuous).

Lemma 2: If $X$ is contractible, then all homotopy groups are trivial.

Proof: Given a map $g:S^n\rightarrow X$, we need to find a homotopy between $g$ and a constant map. Well, let $F$ be a homotopy witnessing the contractibility of $X$. Then the map $G:S^n\times I$ defined by $G(x,t) = F(g(x),t)$ has the desired properties.

Putting this altogether shows that $\pi_i(\mathbb{Z}) = \pi_i(\mathbb{R}) = \{e\}$ for $i > 0$ (where we pick a component of $\mathbb{Z}$ once and for all). Can you take it from there?

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Just as a remark - and in the end it amounts to Jason DeVito's answer - but you don't need the full power of the long exact sequence of a fibration to prove this - just general covering space theory.

In fact it is a general fact (which does however easily follow from the LES) is that for $n \ge 2$ a covering $\tilde{X} \to X$ induces an isomoprhism $\pi_n(\tilde{X}) \to \pi_n(X)$. The point is you don't need to think about fibrations if you aren't familiar with them (see Prop 4.11 in Hatcher for details).

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You can lift the sphere map $f: S^k \to S^1$ through the universal covering space $\pi: \mathbb R\to S^1$ since $1=f_*\pi_1(S^k)\leq\pi_1(S^1)=\mathbb Z$ and contract inside $\mathbb R$ for $k\geq 2$. I am not sure you can do HES through disconnected spaces.

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