Showing that $\pi_2 (\mathbb S^2)$ is infinite using homotopy pointed class of maps $\mathbb S^2 \to \mathbb S^2$ I'd like to understand the proof of $\pi_n(\mathbb{S}^n)\simeq \mathbb{Z}$ if $n \geq 1$, (related Suspension homomorphism)
The proof is "linear" but I don't understand the key part :
Thanks to Freudenthal's theorem we have that $\pi_{n}(\mathbb{S}^n) \simeq \pi_{n+1}(\mathbb{S}^{n+1})$ if $n \geq 2$ so in order to prove the thesis, is sufficient to show that $\pi_2(\mathbb{S}^2)$ is infinite, since again by Fredenthal $\pi_1({\mathbb{S}^1}) \longmapsto \pi_2(\mathbb{S}^2)$ is a surjective homomorphism.
To accomplish this, in my notes the following fact is used : "we use that we can associated to homotopy pointed classes of maps from $\mathbb{S}^2$ to $\mathbb{S}^2$, i.e $[\mathbb{S}^2,\mathbb{S}^2]^0$ an homomorphism in $\text{Hom}(H_2(\mathbb{S}^2))\longmapsto \text{Hom}(H_2(\mathbb{S}^2))$.
Apparently this homomorphism is surjective since if we think $\mathbb{S}^2 = \mathbb{C} \cup \lbrace \infty \rbrace \longmapsto \mathbb{C} \cup \lbrace \infty \rbrace$ sending $z \to z^n$ and $\lbrace \infty \rbrace \to \lbrace \infty \rbrace$ this induces the map $H_2(\mathbb{S}^2) \longmapsto H_2(\mathbb{S}^2)$ such that $1 \to n$.
I don't understand how to prove that the "connecting" map from $[\mathbb{S}^2,\mathbb{S}^2]^0$ to $\text{Hom}(H_2(\mathbb{S}^2),H_2(\mathbb{S}^2))$ is an homomorphism and how to see that the map sending $z \to z^n$ and $\lbrace \infty \rbrace \to \lbrace \infty \rbrace$ induces the multiplication by $n$.
Any help,hint or reference would be appreciated.
 A: You don’t actually need to use that the “connecting map” $[S^2,S^2]^0 \rightarrow \mathrm{Hom}(H_2(S^2),H_2(S^2))$ to be a homomorphism. You just need to show that it has infinite image.
Now, we can replace $S^2$ with $\mathbb{CP}^1=\mathbb{C} \cup \{\infty\}$. Let $f: z \longmapsto z^n$. We can write $\mathbb{CP}^1=U \cup V$ with $U=\{|z| \leq 1\}$, $V=\{|z|>0\} \cup \{\infty\}$, and thus use the Mayer-Vietoris sequence for the action of $f$ on $0=H_2(U\cap V) \rightarrow 0=H_2(U) \oplus H_2(V) \rightarrow H_2(U\cup V) \rightarrow H_1(U \cap V) \rightarrow 0=H_1(U) \oplus H_1(V)$.
This works because $U \cap V$ is homotopic to $S^1$ and $U,V$ are contractible.
But let $g: S^1 \rightarrow S^1$ be given by $z \rightarrow z^n$, then we have an isomorphism $H_1(S^1) \rightarrow H_1(U \cap V)$ where $f$ acts on the right and $g$ on the left. But $g$ acts by multiplication by $n$, so that $f$ acts on $H_1(U\cap V)$ by multiplication by $n$. So $f$ acts on $H_2(S_2) = H_2(U \cup V) \cong H_1(U \cap V)$ by multiplication by $n$.
