TITLE: Ring homomorphisms which map a unit to a unit map unity to unity?
QUESTION [8 upvotes]: this is the third part of a question I've been working on from Hungerford's Algebra. It is exercise 15 in the first section of Chapter III.

$(c)$ If $f\colon R\to S$ is a homomorphism of rings with identity and $u$ is a unit in $R$ such that $f(u)$ is a unit in $S$, then $f(1_R)=1_S$ and $f(u^{-1})=f(u)^{-1}$. 

I see how $f(u^{-1})=f(u)^{-1}$ follows from $f(1_R)=1_S$, for if that is so, then 
$$f(1_R)=f(uu^{-1})=f(u^{-1}u)=1_S\implies f(u)f(u^{-1})=f(u^{-1})f(u)=1_S.
$$
It seems easy, but I can't manage to show $f(1_R)=1_S$. I was hoping that someone could perhaps give me a hint on how to proceed on showing the first part? Thank you.

REPLY [6 votes]: You can cancel $f(u)$ in the equation $f(u)=f(1_R)f(u)$.