Is there a homomorphism from a full product of finite cyclic groups onto $\mathbb Z$? Trying to answer this question, I encountered the following question, the answer to which should be known but it is hard to Google, so I did not find it. 
Let $G=\prod_{n\in\mathbb N}\mathbb Z_n$ be a full product of finite cyclic groups $\mathbb Z_n=\mathbb Z/n\mathbb Z$. Exists there a homomorphism $f:G\to\mathbb Z$ such that $f((1,1,\dots))=1$?
According to this question, my question should be equivalent to the following.
Is the group $\{(m,m,\dots)\in G: m\in\mathbb Z\}$ a direct summand of the group $G$? 

The only homomorphisms $\varphi:\prod_{n \in \mathbb{N}} \mathbb{Z} \to \mathbb{Z}$ are the obvious ones
$$\varphi\left(a_1,a_2,\dots\right)=\sum_{n\in\mathbb{N}}a_nb_n,$$ 
where $\left(b_1,b_2,\dots\right)$ is a sequence of integers with $b_n=0$ for all but finitely many $n$. In particular, $\varphi$ is determined by its restriction to $\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$. For a proof, see for example https://mathoverflow.net/questions/10239/is-it-true-that-as-z-modules-the-polynomial-ring-and-the-power-series-ring-ove/10249#10249.
But the obvious map $\theta:\prod_{n \in \mathbb{N}} \mathbb{Z} \to \prod_{n \in \mathbb{N}} \mathbb{Z}/n\mathbb{Z}$ sends every element of $\bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ to a torsion element, which must therefore be in the kernel of any homomorphism  $\alpha:\prod_{n \in\mathbb{N}} \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}$. So by the theorem above, $\alpha\circ\theta=0$ and so $\alpha=0$. 
