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The question says:

Theorems of Hurewicz and Hopf say that for $k < n, \pi_{k}(S^n)=1$ and $\pi_{n}(S^n)\cong \mathbb{Z}$. Assuming this for the moment, use the Hopf fibration $\eta : S^3 \rightarrow S^2$ with fibre $S^1$ to calculate $\pi_{3} (S^2).$

My question is:

How is the answer of this question different from the answer of the one in the following link Hopf fibration and $\pi_3(\mathbb{S}^2)$? I feel like my question is much easier, could anyone give me a hint and an outline for the solution please?

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    Do you know anything about the long exact sequence of homotopy groups induced by a fibration? In particular, you have an exact sequence $${1}\cong \pi_3(S^1)\to \pi_3(S^3)\to \pi_3(S^2)\to \pi_2(S^1)\cong {1}.$$ – Batominovski Oct 03 '19 at 14:22
  • yes I know .... but I do not know the link between my question and them ..... could you explain this for me please?@WETutorialSchool –  Oct 03 '19 at 22:15

1 Answers1

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Let $F\to E\to B$ be a fibration. Then we have the following long exact sequence of homotopy groups $$\ldots \to \pi_{n+1}(B)\to \pi_n(F)\to \pi_n(E)\to \pi_n(B)\to \pi_{n-1}(F)\to\ldots.$$ In particular, for the Hopf fibration $S^1\to S^3\to S^2$, we obtain $$\ldots \to \pi_{n}(S^1)\to \pi_n(S^3)\to \pi_n(S^2)\to \pi_{n-1}(S^2)\to \ldots.\ \ \ \ \ (1)$$

We know that $\pi_n(S^1)=\Bbb Z$ for $n=1$ and $\pi_n(S^1)=\{1\}$ for $n>1$. Hence when $n\ge 3$, (1) looks like $$\ldots\to \{1\} \to \pi_n(S^3)\to \pi_n(S^2) \to \{1\}\to\ldots.$$ Since the sequence is exact, we conclude that $$\pi_n(S^3)\cong \pi_n(S^2)\ \ \ \ \ (2)$$ for all $n\ge 3$.

Now the Hurewicz theorem gives us $\pi_3(S^3)=\Bbb Z$. So (2) means that $$\pi_3(S^2)\cong \Bbb Z.$$

Batominovski
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  • Could you look at that question, please?https://math.stackexchange.com/questions/3379122/show-that-pi-nb-is-isomorphic-to-pi-ne – Intuition Oct 04 '19 at 07:41
  • Could you please tell me what theorem in Allen Hatcher that tell me if I have a vibration then I have the mentioned long exact sequence? –  Oct 04 '19 at 16:56
  • From where did you get this information $\pi_n(S^1)={1}$ for $n>1$. ? –  Oct 04 '19 at 17:35
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    To answer your first question, you can find this in Theorem 4.41 on page 376 of Hatcher's book (https://pi.math.cornell.edu/~hatcher/AT/AT.pdf). – Batominovski Oct 04 '19 at 17:40
  • To answer your second question, there is a fibration $\Bbb Z\to \Bbb R\to S^1$ (which?). Then, use the same argument as above, we have the exact sequence $$\ldots \to \pi_n(\Bbb Z)\to \pi_n(\Bbb R) \to \pi_n(S^1) \to \pi_{n-1}(\Bbb Z) \to \ldots.\ \ \ \ \ (3)$$ Because $\Bbb Z$ is a discrete topological space, $\pi_n(\Bbb Z)={1}$ for all positive integers $n$, and because $\Bbb R$ is contractible, $\pi_n(\Bbb R)={1}$ for all positive integers $n$. Thus (3) looks like $$\ldots\to{1}\to {1} \to \pi_n(S^1) \to {1}\to\ldots$$ for $n>1$. The rest is all yours. – Batominovski Oct 04 '19 at 17:43
  • this fibration occurs when $n=1$ .... correct? –  Oct 04 '19 at 17:47
  • but still I do not understand how your answer answers my question –  Oct 04 '19 at 17:48
  • Sorry I got it :)) thanks! –  Oct 04 '19 at 17:51
  • Hmm?? The fibration has nothing to do with $n$ (presumably, you are referring to $n$ in $\pi_n$). – Batominovski Oct 04 '19 at 17:51
  • frankly know which fibre is this. –  Oct 04 '19 at 18:07
  • You mean you don't know which fibre $\Bbb Z\to \Bbb R\to S^1$ I was referring to? It is given by $p:\Bbb R\to S^1$ sending $t\mapsto e^{2\pi t}$. – Batominovski Oct 04 '19 at 18:13