Let $G$ be a group and $N\triangleleft G$ be a normal subgroup. Then $G\cong N\rtimes H$ for some subgroup $H\leq G$ if and only if there exists a group homomorphism $\psi:G/N\rightarrow G~,~\psi(gN)\in gN$ for all $g\in G$.
My best attempt is as follows.
$(\Rightarrow)$: I am not very sure how to define the group homomorphism $\psi$ to obtain the desired result.
$(\Leftarrow)$: Suppose that we have a group homomorphism $\psi:G/N\rightarrow G~,~\psi(gN)\in gN$ for all $g\in G$. Let $H\leq G$ be the subgroup such that ${\rm ker}\left(\psi\right)=H/N\leq G/N$. Define $\phi:N\rtimes_{f} H\rightarrow G$ such that $(n,h)\mapsto nh$, where $f:H\rightarrow {\rm Aut}(N)~,~h\mapsto f(h)(n)=hnh^{-1}$. Then \begin{align*} \phi\left(\left(n_{1},h_{1}\right)\left(n_{2},h_{2}\right)\right)&=\phi\left(n_{1}f(h_{1})(n_{2}),h_{1}h_{2}\right) \\ &=n_{1}f(h_{1})(n_{2})h_{1}h_{2} \\ &=n_{1}h_{1}n_{2}h_{2} \\ &=\phi\left(n_{1},h_{1}\right)\psi\left(n_{2},h_{2}\right) \end{align*} so that $\phi$ is a group homomorphism. $\phi$ is 1-1 since if $(n,h)\in{\rm ker}(\phi)$, then $nh=\phi(n,h)=1_{G}$ which gives $n=h^{-1}$. Note that $H$ is a subgroup containing $N$ and so $N\cap H=N$. Thus, $\psi(h^{-1}N)\in N$ and so $n_{0}=\psi(h^{-1}N)$ for some $n_{0}\in N$.
I would appreciate any hints, preferably without the use of exact sequences. Thank you.