# cohomology of infinite product of EM spaces

Is it true that any map from an infinite product of Eilenberg-MacLane spaces in to an Eilenberg-MacLane space factor through, upto homotopy, a finite subproduct? Even if we take with field coefficients?

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I don't think so. Let $X$ be a product of infinitely many circles, then $\pi_1(X)$ is the product of infinitely many copies of $\mathbb Z$. Know take the map $X \to K(\pi_1(X),1)$. This does not factor through any finite subset of the circles because if it did the map would factor through a finitely generated group. –  Will Sawin Mar 23 at 5:54

This is not true because it's not true in algebra. Any map $A \to B$ of abelian groups can be realized by a map $K(A,n) \to K(B,n)$ which is unique up to homotopy, and so we can just find a map $\Pi A_i \to B$ which does not factor through a finite subproduct.
If $\mathbb F$ is a field, we can construct a map $\Pi^\infty \mathbb F \to \mathbb F$ which does not factor through any finite subproduct using the Axiom of Choice. Let $e_i$ be the element in the product which is 1 in the i'th position and 0 elsewhere. These are linearly independent, and so we can extend them to a basis of $\Pi^\infty \mathbb{F}$. Let $f$ be the unique linear function that sends all the $e_i$ to 1 and sends the rest of the basis elements to 0.
Even more interesting is that we can take $\prod_{p \text{ prime}} \mathbb{Z}/p$. Any finite subproduct is a finite group, but this infinite product has nontorsion elements (e.g. $(1,1,1,1\ldots)$ is nontorsion) and so there are nontrivial maps from this infinite product to $\mathbb{Q}$.
This is certainly the "right" answer, but it is not strictly speaking correct for the question as stated. It is typically not true that an infinite product of $K(A_i,n)$s will have the homotopy type of a CW-complex, so maps from the product to $K(B,n)$ will not necessarily correspond to the group maps. In fact, for $n=0$, it is true that any map from a product of discrete spaces to another discrete space factors through a finite subproduct. –  Eric Wofsey Mar 23 at 16:20