I believe it is known that if I is a set of non-measurable cardinality, then any homomorphism $Z^I\to Z$ factors through a finite power. Here $Z$ is the group of integers. Can anyone give a reference for this?



Homomorphisms from powers of Z to Z
This runs under the name **Łoś-Eda Theorem**. A reference is the book Paul C. Eklof, Alan H. Mekler, *Almost Free Modules* (2002): 


Call a set $I$ *$\omega$-measurable* if its cardinality is greater or equal to the first measurable cardinal. This is equivalent to $I$ being uncountable and supporting a non-principal countably complete ultrafilter. 


First note that $\mathbb{Z}$ is slender (Cor. III.2.4). Then, by Cor. III.3.6 (and the discussion before Lemma III.3.5), if $I$ is not $\omega$-measurable, the natural map 
$$\phi: \bigoplus\_{i \in I}\operatorname{Hom}(\mathbb{Z},\mathbb{Z}) \to \operatorname{Hom}(\prod\_{i \in I}\mathbb{Z},\mathbb{Z}),\; (g\_i)\_i \mapsto \big((m\_i)\_i \mapsto \sum\_i g\_i(m\_i)\big)$$
is an isomorphism. 


**Remarks:** 1) If $I$ is $\omega$-measurable, not all homomorphisms $\prod\_I \mathbb{Z} \to \mathbb{Z}$ factor through a finite subset of $I$. For, let $D$ be a non-principal countably complete ultrafilter on $I$ and let $K\_D = \lbrace x \in \prod\_I \mathbb{Z} \mid I \setminus \sup(x) \in D\rbrace$. Then it's not hard to show that the composition $\prod\_I \mathbb{Z} \twoheadrightarrow \prod\_I \mathbb{Z}/K\_D \cong \mathbb{Z}$ doesn't factor through a finite subset of $I$ (the latter isomorphism uses II.3.3). 


2) Irrespective whether $I$ is $\omega$-measurable or not, there is a canonical isomorphism 
$$\operatorname{Hom}(\prod\_{i \in I}\mathbb{Z},\mathbb{Z}) \cong \bigoplus\_D \operatorname{Hom}(\mathbb{Z},\mathbb{Z})$$
where $D$ runs through all countably complete ultrafilters on $I$ (Cor. III.3.7). 


