graph theory
Given a strongly connected directed graph $G=(V,E)$, and a node $r \in V$. Let $T_r$ be the set of spanning trees of $G$ with $r$ as root and all edges pointing to $r$. Is is it possible that there is an edge $e$ such that for all $t \in T_r$, we have $e \in E(Tr)$? In other words, is it possible that there is a directed edge of $G$ belongs to all possible spanning trees with $r$ as root?
BTW : My guess is that it is impossible.
Edit : I meant to say "directed graph"
It depends on how you see undirected edges in presence of directed edges. Depending on your need, you can have your own definition of 'strongly connected' and define it accordingly.
As far as I know, if one says 'directed graph' then one usually means that all edges are directed. And if a graph is not directed, then it is undirected.
Here, you can also treat undirected edges as 'bi-directed' edges i.e. you can traverse in any direction on these edges. If you see undirected edges this way then yes, you can call a graph which has at least one directed edge, a 'directed graph'.
It is easy to see that having indegree, outdegree $\ge 1$ for every vertex is a necessary condition for a graph to be strongly connected. Otherwise, you would have an unreachable vertex or a vertex from which there are no paths.
But it is not a sufficient condition; counterexample below.
Draw two directed triangle graphs and join them by a single directed edge.
Since the degrees are not balanced, there is a directed path from one strongly connected component to the other, but not the other way around; despite the underlying undirected graph being connected.
Best Answer
Unless I misunderstand something, of course it is possible. There will always be such an edge if the graph can be divided into two parts with only two edges (one in each way) between them.
E.g. the full simple graph on two vertices. There is only one spanning tree starting at each vertex and it has one edge.