Proof: directed cycle?

Question

A directed cycle graph is a directed version of a cycle graph, with all the edges being oriented in the same direction.

In a directed graph, a set of edges which contains at least one edge (or arc) from each directed cycle is called a feedback arc set. Similarly, a set of vertices containing at least one vertex from each directed cycle is called a feedback vertex set.

A directed cycle graph has uniform in-degree 1 and uniform out-degree 1.

Directed cycle graphs are Cayley graphs for cyclic groups (see e.g. Trevisan).

Answer 1

Hint: Suppose that when you removed the edge, still remains just a connected component. Take the two vertices that were joint by that edge. What happens with those two vertices? Why there is going to be a contradiction?

Answer 2

Consider a tree $T$ and assume that $T-e$ is connected where $e=xy \in V(T)$. Since $T-e$ is connected we know that there exists a $u-v$ path in $T-e$ for all $u,v \in V(T-e)$. Let $w\in V(T-e)$. We know that there exists a $w-x$ path and a $w-y$ path in $T-e$. Adding the edge $e$ produces a cycle $C$ in T which is a contradiction since $T$ is a tree (acyclic). Thus removing an edge from a tree results in a disconnected graph with two components, that is, every edge in a tree is a bridge.