I'm not very familiar with the classical theorems and results that are well known to experts in graph theory, hope you don't mind if my question is too naive.
Given bi-regular bipartite graph $G=(V,E)$, where $V=V_1 \bigcup V_2$ and $V_1 \bigcap V_2=\emptyset$. Suppose $\text{deg}(u)=m, \forall u\in V_1$ and $\text{deg}(v)=n, \forall v\in V_2$. It then holds that $m|V_1|=n|V_2|$.
The question is: suppose there exists an integer $k < m$ such that $k|V_1|=n L$ where $L < |V_2|$ is an integer, can we always find an induced bi-regular subgraph by carefully choosing $L$ vertices in $|V_2|$ and all vertices in $V_1$?
The above condition is actually the sufficient condition if there exists a bi-regular graph, with $k$ and $n$ being the degrees of vertices.
Thank you for your help!
It is easy to deal with this problem in terms of covers of $V_1$. Namely, for each vertex $v_2\in V_2$ a set $N(v_2)$ of its neighbors is a subset of $V_1$ of size $n$. Each vertex $v_1\in V_1$ is covered by a family $\mathcal U=\{N(v):v_2\in V_2\}$ exactly $m$ times. Given $k<m$ such that $kn$ is divisible by $r=|V_1|$ we want to find a subfamily $\mathcal V$ of a given $n$-uniform cover $\mathcal U$, which covers each vertex of $V_1$ exactly $k$ times. This is not always possible. For instance, recently Andriy Yurchyshyn, solving an other problem, for $r\in\{6,10,14,18\}$ constructed a family $\mathcal U$ of subsets of a set $[r]=\{1,\dots,r\}$ of size $n=r/2$ each, such that each element $v_1\in [r]$ is covered exactly $m=\tfrac 14 {r \choose r/2}$ times and $\mathcal U$ contains no subfamily $\mathcal V$ which covers each element $v_1\in [r]$ exactly $k=1$ times. The family $\mathcal U$ for $r=6$ is
The following subfamilies of $\mathcal U$ also produce counterexamples:
and