How to prove that $Ф(1) = 1'$ if $R'$ is an integral domain?

Question

Since $R'$ is an integral domain , $Ф(b) = Ф(b.1) = Ф(b).Ф(1) = Ф(1).Ф(b)$ .But i can prove this only for those $r'∈R'$ for which there $∃r∈R$ such that $Ф(r)=r'$.

Answer 1

You can't prove it, because there's a small mistake: we can't rule out the possibility that $\phi$ is the map that sends everything to $0$.

So if $R'$ is an integral domain, then we can show that either $\phi(1)=0$ or $\phi(1)$ is an identity of $R'$ as follows: $\phi(1)=\phi(1^2)=\phi(1)^2$, so if we assume that $\phi(1) \neq 0$, then we get for every $r' \in R'$, $\phi(1)(r'-\phi(1)r')=0$, so $r'=\phi(1)r'$, thus $\phi(1)$ is an (or the) identity element in $R'$.

Answer 2

Hint: $\phi(1)$ is idempotent.