Prove that D is strongly connected if and only if for every edge-cut S of the underlying graph G that separates V (G − S) into two sets A and B I'm struggling with writing a detailed proof for this.
Let D be a digraph with at least two nodes, and let G be the underlying graph of D.
An edge-cut of G is a set S of edges of G such that G−S is disconnected, where G−S
denotes the graph obtained from G by deleting all edges in S.
Now,
Prove that D is strongly connected if and only if
for every edge-cut S of the underlying graph G of D that separates V (G − S) into two sets A and B,
there is an arc in D directed from a node in A to a node in B and an arc in D directed from a node
in B to a node in A.
So far, i have assumed D is strongly connected.
Taken u and v as two vertices in D, and since D is strongly connected both vertices are reachable from one another (path from u to v and a path from v to u).
Then an edge cut is performed on G, such that v and u are separated into two connected components, then v $\in$ A and u $\in$ B.
I know that the edges that are cut represent the path from u to v and from v to u, but don't know how to explain via a proof. Also how do i explain the only if part?
 A: For the forward part: Let $D$ be strongly connected connected, and let $S\subseteq E(G)$ be a cut inducing the components with vertex sets $A$ and $B$. Moreover, let $v\in A$ and $u\in B$. Since $D$ is strongly connected, there exists a directed path $P_{vu}$ from $v$ to $u$. Since $u$ and $v$ are in different components of $G-S$, $P_{uv}$ contains an edge in $S$. The argument is identical for proving there is an arc from $B$ to $A$. Since $S$, $a$ and $b$ were arbitrary, we are done.
Suppose now that for every edge-cut $S$ of $G$ of that separates $V (G − S)$ into two sets $A$ and $B$, there is an arc in $D$ directed from a node in $A$ to a node in $B$ and an arc in $D$ directed from a node in $B$ to a node in $A$. Consider any two vertices $u$ and $v$. We proceed by constructing a directed path from $v$ to $u$. Let $A_1 :=\{v\}$ and let $S_1$ be the set of outgoing edges from $v$, and let $B_1:= V(G)\setminus A_1$. By assumption, there is an edge $e=vv_1$ outgoing from $v$ to a vertex in $B_1$. If $v_1=u$ we are done, by adding $u$ to $A_1$ forming $A_2$, where $A_2$ contains a path from $v$ to $u$. Otherwise, iterate this process of adding vertices from $B_1$ to $A_1$ forming $A_2$ and $B_2$ and so on. By assumption, this can always be done, and eventually this iterative process stops when $v$ is added to $A_i$. At each step of this iteration, we only consider adding vertices which have an incoming edge from a vertex in $A_i$. Thus, there is a path from $v$ to the vertex added at step $i$. Thus, there is a path from $v$ to $u$.
