In a directed graph, a vertex $a$ can reach every vertex, and every vertex can reach another vertex $b$. Can we always sort all the edges as $e\_1,e\_2,\ldots,e\_n$ so that every prefix $e\_1,e\_2,\dots,e\_i$ (when viewed as undirected edges) forms a connected subgraph, and similarly for every suffix $e\_i,e\_{i+1},\dots,e\_n$?


I believe this is true if there is a path from $a$ to $b$ that contains all other vertices. Suppose the path is $a\rightarrow v\_1\rightarrow\dots\rightarrow v\_k\rightarrow b$. We let the edge $a\rightarrow v\_1$ come first, then edges that connect $v\_2$ to the set $\{a,v\_1\}$ (in any order), then edges that connect $v\_3$ to the set $\{a,v\_1,v\_2\}$, and so on. But is the statement true if there is no path from $a$ to $b$ that contains all other vertices?



Vertex reachability in directed graph
Add to $G$ a new edge $e=\vec{ba}$, the new graph $G\cup e$ is strongly connected. Start with a directed cycle containing $e$, let it consist of edges $e\_1,e\_2,e\_3,\dots,e\_m,e$, where $e\_1$ is incident to $a$ and $e\_m$ is incident to $b$. Start with a sequence $e\_1,e\_2,\dots,e\_m$. Now I use an ear decomposition of $G$: the edge set $E$ is a disjoint union of $E\_1=\{e\_1,e\_2,\dots,e\_m,e\}$, $E\_2$, $E\_3$, $\dots,E\_s$, where each $E\_i$, $i=2,3,\dots,s$ is an ear: it is a set of edges of a simple directed path or cycle $v\_1 v\_2\ldots v\_r$ in which all vertices except $v\_1$ and $v\_r$ are not covered by the set $F\_{i-1}:=E\_1\cup\ldots\cup E\_{i-1}$ (while $v\_1$ and $v\_r$ are covered by $F\_{i-1}$). The existence of an ear decomposition in a connected directed graph which starts from a given cycle is well known and easy to prove. So your conjecture follows by induction from the following 


**Lemma.** Let $H$ be an undirected graph in which the edges are enumerated by $e\_1,\ldots,e\_t$ so that for all $i=1,2,\ldots,t$ the graphs formed by $e\_1,\ldots,e\_i$ and $e\_i,\ldots,e\_t$ are connected. Assume also that there are vertices $a,b\in V(H)$ such that $a\in e\_1,b\in e\_t$ and all other vertices of $H$ have degree at least 2. Let $v\_1\ldots v\_r$ be an ear for $H$, that is, a simple path or cycle with $v\_1,v\_r\in V(H)$ and $v\_2,\ldots,v\_{r-1}$ being new vertices. Then the graph $H$ plus this ear also has such an enumeration of edges.


**Proof.** If $\{v\_1,v\_r\}\cap \{a,b\}\ne \emptyset$, say $v\_r=b$, add the edges $v\_1v\_2,v\_2v\_3,\dots,v\_{r-1}v\_r$ after $e\_t$. Otherwise $v\_1,v\_r$ are not lists of $H$ and each of them is covered at least by two edges.
Choose the minimal $i$ for which $e\_i$ covers either $v\_1$ or $v\_r$. Without loss of generality $v\_1\in e\_i$. Then $i<t$ and $v\_r$ is covered by $e\_{i+1},\dots,e\_t$. Insert the edges $v\_1v\_2,v\_2v\_3,\dots,v\_{r-1}v\_r$ between $e\_i$ and $e\_{i+1}$.


