TITLE: Bijection between perfect matchings permutations with even cycles
QUESTION [11 upvotes]: It is possible to proof that the number of perfect matching on a set of $2n$ elements is $n!!$, and on the other hand, it is also possible to proof that the number of permutations $\varphi$ of a set of $2n$ elements with the property that every cycle of $\varphi$ has even length is $(n!!)^2$. This implies that there exists a bijection between the aforementioned set and the pair of perfect matching. I wonder if someone has a nice interpretation for this bijection; meaning, how can we prove this constructing the bijection directly. Thanks!

REPLY [2 votes]: A simple induction on $n$ will prove the result. For $n=2$, $n! = 2$ perfect matchings. Now suppose, for $n=r \ge 2$, there are $r!$ perfect matchings in a (complete) bipartite graph on $2r$ nodes. 
Then find the number of perfect matchings in a bipartite graph that you construct on $2(r+1)$ nodes.
Also, by contradiction-- you can find a shorter proof.
PS: I suppose you meant the permutation of a set of $n$ elements, not $2n$.
The Bijection $f: M(G) \rightarrow S_n$:
Let $G = (V\cup W, E)$ be a bipartite graph, where   $V = \{v_1, v_{2}, \cdots, v_n\}$,
  $W = \{w_1, w_{2}, \cdots, w_n\}$, and $E \subseteq V \times W$. Here $M(G)$ is the set of perfect matchings in $G$.
Then $\forall m \in M(G),\, \exists$ $\pi \in S_n$, such that
 $\forall (v_i, w_j) \in m \Longrightarrow i^\pi = j$.
PS2: The number of cycles can never be larger than $O(n\cdot n!)$.