Suppose $\Phi:G_1\to G_2$ is an epimorphism from a finite group $G_1$ to $G_2$. Assume $G_1$ can be generated by m elements and let $\{z_1,z_2,\ldots,z_m\}$ generate $G_2$. Then. can we say there exists a set of elements $\{g_1,g_2,\ldots,g_m\}$ in $G_1$ which generate $G_1$ and $\Phi(g_i)=z_i$ for every $i=1,2,\ldots,m$? If it holds, how can we say that the pre-images of $z_i$'s, where $i=1,2,\ldots,m$ will generate group $G_1$? Is there any direct way to prove it?
What I have tried is:
$\Phi$ is a surjective map, so we take a set of preimages of $\{z_i:i= 1,2,\ldots,m\}$, say $\{\bar g_1,\bar g_2,\ldots,\bar g_m\}$ in $G_1$. Suppose $K=\ker\Phi$. Here $\Phi$ need not be injective. Now, we consider a set $\{\bar g_1K,\bar g_2K,\ldots,\bar g_mK\}$. If we can show that this set generates the quotient group $G_1/K$ (I don't know how to prove it...), then there exists a generating set $\{g_1,g_2,\ldots,g_m\}$ of $G_1$ such that $g_i\in\bar g_iK$ (I have used a result from a German paper which states: "Let $G$ be a group generated by $d$ elements. If $N$ is a finite normal subgroup, then, for every generating set $\{\bar e_1,\bar e_2,\ldots,\bar e_d\}$ of $G/N$, there exists a generating set $\{e_1,...,e_d\}$ of $G$ such that $e_i\in\bar e_iN$." Here, I am providing a link to that paper. And this gives $\Phi(g_i)=z_i$ for every $i=1,2,\ldots,m$.
Is this approach correct? and if it is then how to prove the set $\{\bar g_{1}K,\bar g_{2}K,...,\bar g_{m}K\}$ generates $G_{1}/K$ or any hint ?
Thank you.