Is there a counter example or proof for the claim that the lightest edge-disjoint union of a pair of perfect matchings contains the edges of the lightest perfect matching in a finite complete graph with $2n$ vertices and positive edge-weights?
Make the black edges cost 1, green edges cost 2, and purple edges cost 1 billion. Make all other edges cost 10 trillion so we can safely ignore them (thus this counter example also works for the version of the question concerning complete bipartite graphs).
The minimum perfect matching is the three black edges, but any edge-disjoint union of two perfect matchings that contains the black edge perfect matching also must contain at least one purple edge. Whereas you can for instance instead get an edge-disjoint union of two perfect matchings that's the 4 green edges plus the bottom and top black edges.