Let $R_1,R_2$ be two rings and suppose that $f:R_1\to R_2$ is a ring homomorphism. Denote $R_1'$ and $R_2'$ for the groups of units of $R_1$ and $R_2$. Next, let $f':R_1'\to R_2'$ be the restriction of $f$ to $R_1'$. I've already proven that this is well-defined and that $f'$ is a group homomorphism. Finally, suppose that $f'$ is surjective. Does this imply that $f$ is surjective?
So far I've tried a couple of things. First I tried to prove this, but couldn't think about anything that helped. I can't really see any connection between units and other elements from some arbitrary ring, so I started trying to come up with some counter examples. This gave me the idea to use $R_2=\mathbb{Z}$, since it is has a very small set of units compared to its size. Next I was trying some finite rings with more than 2 units as candidates for $R_1$, but couldn't really think of any ring homomorphism that would make sense. I've mostly tried rings of the form $\mathbb{Z}/n\mathbb{Z}$ for some integer $n$, all without success. Am I forgetting something obvious? Or is it actually true?
Best Answer
The units in $R=\Bbb Z[X]$ are $\pm1$. The ring homomorphism $R\to R$, $p(X)\mapsto p(X^2)$ is not onto.