TITLE: Does there exist an onto group homomorphism from $\mathbb{R}\to \mathbb{Z}$?
QUESTION [5 upvotes]: Consider $\mathbb{R}$ and $\mathbb{Z}$ as groups with operation as an usual addition. Then does there exist an onto group homomorphism from $\mathbb{R}\to \mathbb{Z}$ ?
My intuition tells me that such homomorphism doesn't exist. So far , I have tried the following.
Suppose $f$ is such homomorphism. Then ,by first isomorphism theorem , $\mathbb{R}/Ker f \cong \mathbb{Z}$. I want to show that the above supposition is wrong. so I have to prove either $\mathbb{R}/Ker f$ is uncountable or it is noncyclic. But I don't know how to go further to prove any cases.

REPLY [6 votes]: Suppose $\phi: \mathbb{R} \to \mathbb{Z}$ is a group homomorphism, and let $x \in \mathbb{R}$, and let $n = \phi(x)$. Then
\begin{align*}
n
&= \phi(x) \\
&= \phi\left((n+1)\left(\frac{x}{n+1}\right)\right) \\
&= (n+1) \phi\left( \frac{x}{n+1} \right).
\end{align*}
Therefore, $\phi\left( \frac{x}{n+1} \right) = \frac{n}{n+1}$.
But this has to be an integer, so we conclude that $\boldsymbol{n=0}$.
(The argument was technically incorrect for $n=-1$. We could use $n^2 + 1$ instead of $n+1$.)
In other words, $\phi(x) = 0$ for all $x \in \mathbb{R}$, since $x$ was arbitrary.