Show that $f^{-1}\{f(a)\}=a\text{Ker}f$

Question

Let $G,G'$ be groups and assume $f: \space G \longrightarrow G'$ is a homomorphism. Show that $\forall f(a) \in \ \operatorname{Im}f$: $$ f^{-1}\{f(a)\}=a \operatorname{Ker}f $$

To begin with, the kernel of the homomorphism is a normal subgroup of $G$, $ \operatorname{Ker}f \triangleleft G.$ This implies that the left and right cosets coincide. Thus, the set of cosets of $ \operatorname{Ker} f$ is: $$ G/ \operatorname{ker}f:=\left\{ a \operatorname{Ker}f, \space \forall a \in G \right\} $$ Now, since $ \operatorname{ker}f \triangleleft G$, $G/ \operatorname{Ker}f$ is a group. Moreover, the fundamental theorem of homomorphisms gives: $$ G/ \operatorname{Ker}f \cong \operatorname{Im}f $$ and so there exists a bijection $\varphi: \space G/ \operatorname{Ker}f \longrightarrow \operatorname{Im}f\;$ s.t. $$ \varphi(a \operatorname{Ker}f)= \operatorname{Im}f, \space \forall a \in G $$ Given that, we must show: $$ f^{-1}[\varphi(a \operatorname{Ker}f)]=a \operatorname{Ker}f $$

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Question:

Which argument about the relation of $f$ and $\varphi$ should one use to prove the initial statement?

Answer 1

You're overthinking the problem.

Just prove that $f(b)=f(a)$ iff $b=ak$ with $k \in \ker f$.