How to prove the product formula for higher homotopy groups?

Question

When I review Hatcher's proof on the fact $$\pi_{n}(\prod X_{\alpha})=\prod_{\alpha}\pi_{n}(X_{\alpha})$$I found I cannot really follow. He wrote "A map $f:Y\rightarrow \prod X_{\alpha}$ is the same thing as a collection of maps $f_{\alpha}:Y\rightarrow X_{\alpha}$. Taking $Y$ to be $\mathbb{S}^{n}$ and $\mathbb{S}^{n}\times I$ gives the result. "

I am confused because the result is intuitive and (probably) trivial, but I do not see how his first line and second line connected together. Presumably the second line means something induced from the earlier proposition on covering maps, but still I do not see how to use this to prove the above statement.

Answer 1

For any map $\mathbb{S}^{n}$ to $\prod X_{\alpha}$, composition with projection give us $f_{\alpha}\rightarrow \prod X_{\alpha}$. So we have a map $$F:\pi_{n}\prod X_{\alpha}\rightarrow \prod \pi_{n}(X_{\alpha})$$

On the other hand given a family of maps $f_{\alpha}:\mathbb{S}^{n}\rightarrow X_{\alpha}$, by other's hint I can form a map $f'：\mathbb{S}^{n}\rightarrow \prod X_{\alpha}$ such that $f'(x)_{\alpha}=f_{\alpha}(x)$. So we have an inverse map $$G:\prod \pi_{n}(X_{\alpha})\rightarrow \pi_{n}\prod X_{\alpha}$$ We have $FG=1, GF=1$. So the two groups are isomorphic.

Alternatively we can to prove $F$ is surjective and injective. Assume $\prod P_{\alpha}\in \prod \pi_{n}(X_{\alpha})$, then $G$ maps it to $\pi_{n}\prod X_{\alpha}$. By $FG=1$ we have $FG(\prod P_{\alpha})$ be $\prod P_{\alpha}$. This showed $F$ is surjective.

Injectivity is also clear because if $p\in \prod \pi_{n}(X_{\alpha})=0$, then there is a null homotopy $I\times \mathbb{S}^{n}$ such that ${0}\times \mathbb{S}^{n}\rightarrow p$, ${1}\times \mathbb{S}^{n}\rightarrow 0$. Then $Gp$ would have a null homotopy as well. This proved the above correspondence is one to one and onto. The fact that $F,G$ are group homomorphisms are directly follow from definition.