This has been put in math.SE for a while without any responses.



> 
> Given $G$ and an $f:V(G)\to{\Bbb N}$, there exists a graph $G\_f$ such that $G$ is $f$-soluble if and only if $G\_f$ has a $1$-factor.
> 
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> 


"G is $f$-soluble" is defined as a graph such that there exists $w:E(G)\to{\Bbb N}$ such that
$$
\sum\_{uv\in E(G)}w(uv)=f(v)
$$
for every $v\in V(G)$.


I believe this is a result from Tutte and it might be related to the $f$-factor theorem. But I don't remember the whole proof. One key step I remember is defining
$$
V(G\_f):=\{v\_1,v\_2,\cdots,v\_{f(v)}|v\in V(G)\}.
$$
Could anyone come up with a proof?



Reduction of $f$-solubility to $1$-factor
Yes, this is a theorem that Tutte used to derive the f-factor theorem from the 1-factor theorem. He had already proved the f-factor theorem directly in 1952, in "The factors of graphs", [MR0048775](http://www.ams.org/mathscinet-getitem?mr=48775). The result you quote is from his paper from two years later, "A short proof of the factor theorem for finite graphs", [MR0063008](http://www.ams.org/mathscinet-getitem?mr=63008). 


If you have trouble locating the reference, I can update this answer to give the actual argument, and construction of what you call $G\_f$.


