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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.

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Forget about the hint and try to prove it directly. When you are done you will notice that you have followed the hint. –  Mariano Suárez-Alvarez Nov 29 '12 at 5:18
    
If the index is finite, then a naive argument on projection into coordinates might suffice. But if the index is uncountable, then I believe I am on the wrong track. Image I need to prove this for $\mathbb{S}^{2}$ and $X=\mathbb{R}\times \mathbb{R}$ graphically, I still could not really follow the hint. Sorry for being slow. –  Bombyx mori Nov 29 '12 at 5:29
    
Why do you say «might»? Have you actually tried to do it? –  Mariano Suárez-Alvarez Nov 29 '12 at 15:02
    
okay. let me construct a proof and you may criticize on it later. –  Bombyx mori Nov 29 '12 at 16:03
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1 Answer

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.

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"I do not know how to form a map..." How about the map $x \mapsto (f_\alpha(x))$? (One can in fact define the product of things as the universal gadget where maps into it are in natural bijection with the collection of maps into each of its factors...) –  Dylan Wilson Nov 29 '12 at 18:10
    
Your argument is not complete. If you really want input on your proof, you should write it out in detail. –  Mariano Suárez-Alvarez Nov 29 '12 at 18:31
    
Let me write in my detail then. –  Bombyx mori Nov 29 '12 at 21:23
    
updated.hopefully it is right this time. –  Bombyx mori Nov 29 '12 at 22:07
    
@MarianoSuárez-Alvarez: Sorry to bother you - can you check my proof? –  Bombyx mori Nov 30 '12 at 3:29
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