Homomorphism property

Question

I have a problem with understanding one of homomorphism properties.

Given $f$ - is a homomorphism, $g \in G$ and $h = f(g)$, then $f^{-1}(h) = g \cdot Kerf$.

Can anyone please prove it?

Answer 1

First you need to know that since $f$ is not necessarily bijective. $f^{-1}(h)$ is not an element, but is the pre-image of $h$ which is a set. So to prove equality of set we need to show that they are subset of each other.

First, let $x\in f^{-1}(h)$, which means $f(x)\in \{h\}$.
Therefore, $f(x)=h=f(g)$.
Write $x=g(g^{-1}x)$.
Since $f(g^{-1}x)=(f(g))^{-1}(f(x))=h^{-1}h=1$, $g^{-1}x\in \ker f$.
Hence $x\in g\cdot \ker f$

Converserly, let $x=ga$ where $a\in \ker f$.
Then $f(x)=f(ga)=f(g)f(a)=f(g)=h$ since $f(a)=1$.
We conclude that $x\in f^{-1}(h)$.

Answer 2

Hint:

Take any element in $x\in g\ker f$, and check $f(x)=h$. This sets $\supseteq$.

For $\subseteq$, take an element $g'$ such that $f(g')=h=f(g)$, and find the value of $f\bigl(g'g^{-1}\bigr)$.