Let $D$ the bipartite digraph with vertices bipartitions $A$ and $B$ such that it has only arcs of $A$ to $B$. It easy to see that $D$ is acyclic (not contains directed cycle).
I know that $D$ has a unique perfect matching $N$ covering the vertices of $A' \cup B'$ with $|A'| = |B'|$, $A' \subseteq A$ and $B' \subseteq B$. I know that we can sort topologically the vertices of $D$ (because is acyclic).
How I can to show that $N$ can be sorted as $(y_1,z_1) , \ldots, (y_t,z_t)$ (Note that $y_i \in A'$ and $z_i \in B'$ for each $i \in \{1, \ldots,t\}$) such that $(y_i,z_j)$ not is arc of $D$ if $i < j$?
In the book of Schrijver, he argument that this is true, because $N$ is a unique perfect matching covering $A' \cup B'$, I not understanded. Anyone know the motive?
Finding the ordering $(y_1, z_1), \dots, (y_t, z_t)$ is a topological sort problem in disguise.
Let $G$ be the following directed graph:
Because $N$ is the unique perfect matching between $A'$ and $B'$, we can show that $G$ is acyclic. If there is a cycle in $G$ with the vertices $(y_1,z_1), \dots, (y_k, z_k)$, then that means that $D$ has all the arcs $(y_2, z_1), \dots, (y_k, z_{k-1}, y_1, z_k)$, and we can use these to replace $(y_1,z_1), \dots, (y_k, z_k)$ and get a different perfect matching between $A'$ and $B'$.
Because $G$ is acyclic, it has a topological sort: we can order its vertices (the arcs of $N$) as $(y_1, z_1), \dots, (y_t, z_t)$ such that when $i<j$, $G$ does not have an arc from $(y_j, z_j)$ back to $(y_i, z_i)$. That is exactly the condition we wanted: that $y_i z_j$ is not an arc of $D$ for any $i<j$.