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?
Best Answer
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 $$\pi_3(S^2)\cong \Bbb Z.$$