Ring Homomorphism Identities

Question

Suppose $f: R\rightarrow S$ is a ring homomorphism such that $R$ is a ring with identity $1_R$. Prove that $f(1_R)$ is the multiplicative identity in $f(R)$

I am stumped on this one. Learning undergrad abstract algebra on my own to prepare.

Any pointers to begin?

Answer 1

Assuming $f$ to be surjective.

Let $s\in S$ then we show $f(1_R)\cdot s = s$. Since $f$ is onto so there exists $r\in R$ such that $f(r)=s$ and so $f(1_R)\cdot s=f(1_R)\cdot f(r)=f(1_R\cdot r)= f(r)=s$ as $f$ is a homomorphism and similarly we can show $s\cdot f(1_R) =s$.

Since $s\in S$ is arbitrary hence $f(1_R)=1_S$. This completes the proof.