Ring homomorphism $\mathbb{Z}/8\mathbb{Z} \to \mathbb{Z}/5\mathbb{Z}$ implies $1=0$ Let $f$ be an ring homomorphism from $R_1$  to $R_2$ and define $f^*$ as the homomorphism from the group of units of $R_1$ to the group of units of $R_2$. Suppose $f^*$ is surjective, the question is to decide if this implies that $f$ is surjective.
After several attempts I got to this example. Take $R_1 =\mathbb{Z}/8\mathbb{Z}$ (so integers mod 8), $R_2 = \mathbb{Z}/5\mathbb{Z}$, and define $f(x) = x \mod 5$ for every $x \in Z/8Z$. 
Now I noticed something weird:
$$1 = f(1) = f(5\cdot 5 \mod 8) = f(5)\cdot f(5) = (5 \mod 5)\cdot (5 \mod 5) = 0\cdot 0 = 0$$
What is wrong with my map? I don't get it, everything seems to be correct. I've checked whether $f$ is a ring homomorphism and it is (altough I'm not so sure anymore). I can't think of other mistake, but there has to be one of course.

The map that you have defined isn't even a homomorphism. In fact, if $m, n$ are coprime, then the only homomorphism between $\mathbb Z/n\mathbb Z$ and $\mathbb Z/m\mathbb Z$ is the trivial one, sending everything to 0: $f(x) = 0$.
