Let $\phi$ be a homomorphism from group $G$ to group $\overline G$. It is to be proven that if $\overline K$ is a normal subgroup of $\overline G$, then $\phi^{-1}(\overline K)=\{k\in G: \phi(k)\in\overline K\}$ is a normal subgroup of $G$.
I tried to prove it as follows:
Suppose on the contrary that $\phi^{-1}(\overline K)$ is not normal to $G$. This means that $\exists x\in G$ such that $xk\ne k'x \;\;\forall k, k'\in \phi^{-1}(\overline K)$. This implies that $xkx^{-1}\notin \phi^{-1}(\overline K)$ for any $k$ in $\phi^{-1}(\overline K) $.$\tag{1}$
But $\phi(xkx^{-1})=\phi(x)\phi(k)(\phi(x))^{-1}\in\overline K$ as $\overline K\trianglelefteq \overline G$, which contradicts $(1)$
Is my proof correct? Thanks for your time.