Question: Let M be the set of maximum matchings of G=(V, E). Find a minimum edge subset R, such that, the intersection of R and M is not empty for each M in M.

Have I understand your question well?

I think this is a good question. Could you send me some background materials?

Let r and m be the cardinalities of R and M respectively. Let M(e) be the set of maximum matchings containing edge e. 

For a complete graph K_n, the question is not hard. For example, r(K_2n) = 2n-1.

Proof. Firstly note that, by symmetry, each edge meets equal number of M's in M. Since K_2n has perfect matchings, each M in M saturates vertex v. Let e_1 ...  e_(2n-1) be all the edges incident with v. It is easily seen that, { M(e_1), ..., M(e_(2n-1) } is a partition of M. Thus, R={e_1 ...  e_(2n-1)} is an edge dominating set. Why R is minimum? Recall that, each edge meets equal number of M's in M.

For K_(2n+1), the question is a little harder, but I think it can be solved by counting its maximum matchings (near-perfect).

Peter Breuer, your dominating set is minimal by inclusion but not by size. For example, in the case of K5, your construction gives 7 edges while the edge set of any subgraph K4 is dominating and contains 6 edges (which is the minimum). However, it is easy to prove that if p=2k+1 then any dominating subset R contains at least 3k = 3(p-1)/2 edges. Indeed, it is known that the edge set of Kp can be partitioned to k Hamiltonian cycles. For each cycle H, we need to take at least 3 of its edges to R, otherwise their exists a maximum matching of Kp formed by the edges of H\R.