$\Lambda$ is a lattice in $\Bbb{C}^g$ and $\Bbb{C}^{g'}$ is a subvector space of $\Bbb{C}^g$ where $\Bbb{C}$ is complex number.
$\textbf{Q:}$ Is $\Lambda$ image in $\frac{\Bbb{C}^g}{\Bbb{C}^{g'}}$ a lattice? I could not see discreteness of $\Lambda$'s image here. If not, what is the counter example?
No. Consider the case where $g=2$, $g'=1$ and consider the lattice $\Lambda$ which is generated as an abelian group by $(1,0),(i,0),(\sqrt2,1),(i\sqrt2,i)$ ($\Lambda$ is a lattice since the elements are a basis for $\Bbb{C}^2$ as a vector space over $\Bbb{R}$). We need to be specific about which subspace of $\Bbb{C}^2$ we are referring to; consider the space $\{(0,x)|x\in \Bbb{C}\}\cong \Bbb{C}^1$. Then in the quotient, $\pi(\Lambda)$ will be generated by $\overline{(1,0)},\overline{(\sqrt2,0)},\overline{(i,0)},\overline{(i\sqrt2,0)}$ (where $\overline{(a,b)}$ denotes a coset in the quotient.). Then $\pi(\Lambda)$ will still be isomorphic to $\Bbb{Z}^4$ as an abelian group. But to be a lattice in the quotient, it needs to be isomorphic to $\Bbb{Z}^2$ since the quotient space has dimension $2$ over $\Bbb{R}$.