I'm looking for a rigorous proof for Hall's bigamy problem:
Let $G$ a bipartite graph with sides $V$ and $U$. Prove that you can match for any vertex in $V$ two vertices in $U$, that are matched only to it if and only if for any $X \subset V$ we have $|N_G(X)| \geq 2|X|$, where $N_G(X)$ is the neighbourhood of $X$ (according to Hall's marriage theorem).
I tried to duplicate the edges in $V$ and apply Hall's marriage theorem, meaning for any $v \in V$ create a $v'$ and connect it to the edges that $v$ is connected to in $U$.
I think that at least one of the sides is sort of trivial but I would love to hear other peoples opinion and get help with the other side or proving both sides if needed. Thanks!
Assume $|N_G(X)|\ge2|X|$ for every $X\subseteq V$. Consider the bipartite graph $G'$ with vertices $V\times\{1,2\}\cup U$ and an edge between $(v,i)$ and $u$ iff $G$ has an edge $vu$. We have the projection $\pi_1\colon V\times\{1,2\}\to V$. Then for $X\subseteq V\times\{1,2\}$, we have $$|N_{G'}(X)|=|N_G(\pi_1(X))|\ge 2|\pi_1(X)|\ge |X|$$ and so by monogamic Hall, there exists a matching of $V\times\{1,2\}$ with $U$ in $G'$. This gives us a bigamic matching of $V$ and $U$ in $G$.
The other direction is trivial.