The title pretty much says it all. Do I have $\pi_n(\Pi_{i=1}^\infty X_i)\cong \bigoplus\limits_{i=1}^\infty \pi_n(X_i)$ or $\pi_n(\Pi_{i=1}^\infty X_i)\cong \Pi_{i=1}^\infty \pi_n(X_i)$?
From what I see in a specific example, it seems that direct sum is right.
Whatever the answer is, I would really like to understand whether there is some sort of categorical (or any other) explanation to it. Intuition tells me that covariant functor $\pi_n$ must map product to product, but I am not $100$% sure whether it is generally true.
Thank you.
Best Answer
We shall show that for any index set $A$ there is a natural bijection
$$i : \pi_n(\prod_{\alpha \in A}X_\alpha) \to \prod_{\alpha \in A}\pi_n(X_\alpha) \tag{1} .$$ $\pi_n(Y) = \pi_n(Y,y_0)$ is the set of pointed homotopy classes $[f]$ of pointed maps $f : (S^n,*) \to (Y,y_0)$. For $n = 0$ this a just a set with a basepoint, for $n = 1$ it has a natural structure of a group, for $n> 1$ a natural structure of a abelian group.
The expression $\oplus_{\alpha \in A}G_\alpha$ only makes sense for abelian groups $G_\alpha$, and it would be weird if you would not have an isomorphism for $n = 0 ,1$ because of the lack of an appropriate object.
Let us define $$i : \pi_n(\prod_{\alpha \in A}X_\alpha) \to \prod_{\alpha \in A}\pi_n(X_\alpha), i([f]) = ([p_\alpha \circ f])_{\alpha \in A} .$$ Here $p_\alpha : \prod_{\alpha \in A}X_\alpha \to X_\alpha$ denotes projection onto the factor with index $\alpha$.
This is clearly well-defined because if $f \simeq g$, then $p_\alpha \circ f \simeq p_\alpha \circ g$ for all $\alpha$. By the universal property of the product each family $(\phi_\alpha)$ of pointed maps $\phi_\alpha :(S^n,*) \to (X_\alpha,x_\alpha)$ has the form $(\phi_\alpha) = (p_\alpha \circ f)$ for a unique pointed map $f : (S^n,*) \to \prod_{\alpha \in A}(X_\alpha,x_\alpha)$, thus $i$ is surjective. To show that $i$ is also injective, note that if $p_\alpha \circ f \simeq p_\alpha \circ g$ for all $\alpha$, then we can choose pointed homotopies $h_n : (S^n,*) \times I \to (X_\alpha,x_\alpha)$ from $p_\alpha \circ f$ to $p_\alpha \circ g$. Again by the universal property of the product we get a pointed homotopy $H : (S^n,*) \times I \to \prod_{\alpha \in A}(X_\alpha,x_\alpha)$ such that $p_\alpha \circ H = h_\alpha$ for all $\alpha$. By construction $H$ is a homotopy from $f$ to $g$. Therefore $i$ is a bijection for all $n$.
For $n > 0$ it remains to be shown that $i$ is a group homomorphism (and thus a group isomorphism). But this is clear because $([p_\alpha \circ f]) =((p_\alpha)_*([f]))$ and each $(p_\alpha)_* : \pi_n(\prod_{\alpha \in A}X_\alpha) \to \pi_n(X_\alpha)$ is a group homomorphism.