Show that $\{x \in R | f(x) = a'\} = a + Ker(f)$ as following.

Question

Let $f:R \to R'$ be a homomorphism and $a\in R, a' \in R'$ such that $f(a)=a'$. Show that $\{x \in R | f(x) = a'\} = a + Ker(f)$.

I have tried find Ker(f) and stuck at $Ker(f) = \{x\in R | f(x) = 0_{R'}\} = \{x \in R | a' = 0_{R'}\}$.

What's next? How to proves above? Any idea? Thanks in advanced.

Answer 1

If $f(x)=a'$, then $$f(x-a)=f(x)-f(a)=0,$$so $x-a\in\ker f$. Conversely, if $x=a+z$ with $f(z)=0$, then $f(x)=a'$.