The question I am trying to answer is as follows. If a digraph has $2n$ vertices and minimum semi-degree (minimum of all in/out degree on each vertex) $n + 1$, must it have a perfect matching consisting of double edges?
What I know is that since each vertex can have a maximum of $2n - 1$ in/out neighbors, there is an overlap of $3$ which means I have at least $3$ double edges on each vertex. For all examples I've tried it turns out to be true, but I'm still not $100$% convinced. I can only really do it with smaller examples before drawing it on paper gets messy.
Does anyone know of a proof for this fact, or if it's not true have any clever ways to come up with a counter-example?
I believe I have a counter-example. Let $n=8$. Consider the following graph:
Using this graph $G$ as a map, we will build a digraph $D$ on $16$ vertices with minimum semi-degree $9$. Let $V(G)=V(D)$ and for each edge of $G$, let that be a double edges of $D$. These will be the only double edges, and since $G$ does not contain a perfect matching, $D$ will not have a perfect double edges matching. Thus all we need to do is construct $D$.
To build the rest of $D$, we will consider the partition of $V(D)$ as $F_1\cup F_2\cup F_3\cup\{v\}$ where $v$ is the central vertex, and each $F_i$ is the vertex set of a component of $G-v$. Add edges going from each vertex of $F_1$ to each vertex of $F_2$, also going from each vertex of $F_2$ to each vertex of $F_3$ and edges from each vertex of $F_3$ to each vertex of $F_1$. This contributes $5$ to the in degree and out degree of each vertex in $V(D)\setminus\{v\}$. In addition, the double edges contribute $3$ to the in-degree and out-degree of each vertex. Thus the vertices in $V(D)\setminus\{v\}$ only need one more edge in and one more out.
Observe that inside $F_1$, there are three edges that do not appear in $G$. Orient these three edges in such a way that the vertex incident with two edges gains both an in-degree and out-degree. This leaves us with one vertex that has the correct semi-degree, two vertices missing one in-degree and two missing one out-degree. Do the same thing for $F_2$ and $F_3$ leaving us with six vertices each needing one in-degree and six vertices each needing one out-degree in $V(D)\setminus \{v\}$. We can that attach the appropriately oriented edges from $v$ to each of these vertices. Then each vertex in $V(D)\setminus\{v\}$ has the correct semi-degree and indeed $v$ also has the correct semi-degree. This completes the construction, giving us a counter-example.