# Non Bridge edges form a cycle in connected graph is equivalent to there exists edge e such that G-e is a tree

We need to show the following are equivalent:

(1) There exists an edge $$e$$ of $$G$$ such that $$G−e$$ is a tree

(2) $$G$$ is connected and the set of edges of $$G$$ which are not bridges form a cycle

This is what I tried:

$$\Rightarrow$$

We know a tree is a graph that is connected and has no cycles. So we have $$G'=G-e$$ is connected & acyclic. Now $$G$$ & $$G'$$ has $$n$$ vertices. We can prove $$G'$$ has $$n-1$$ vertices. Then for $$G'$$ it has $$n$$ vertices & $$n-1$$ edges. Then $$G'$$ is connected. How do I show if edges of $$G$$ which are not bridges form a cycles

Also, I have been able to show $$(1) => G$$ is connected with $$V = E$$. I know a result where this implies $$G$$ is connected & has exactly one cycle. (I have been able to prove it)

$$\Leftarrow$$

$$G$$ is connected and the set of edges of $$G$$ which are not bridges form a cycle. Does this mean $$G$$ is unicycle? As $$G$$ is connected by assumption & it forms a cycle.

$$\Leftarrow$$
All edges of $$G$$ that lie in a cycle are not bridges, because if you remove one of them you can still "take the other part of the cycle to get around". Therefore, $$G$$ has only one cycle $$C$$, and it contains all of the non-bridges. Removing any edge $$e$$ from $$C$$ now destroys the (only) cycle in $$G$$, which gives an acyclic graph. The resulting graph is also still connected because we removed a non-bridge. Therefore $$G-e$$ is a tree.
$$\Rightarrow$$
Removing an edge from a non-connected graph cannot make it connected. Therefore $$G$$ was connected. Since $$G-e$$ is connected, $$e$$ must have been part of a cycle in $$G$$. Since $$G-e$$ is a tree, all cycles of $$G$$ must go through $$e$$ (otherwise they would still be present in $$G-e$$).
Say $$e=\{v,w\}$$. There exists a unique path $$P$$ from $$v$$ to $$w$$ in $$G-e$$, since it is a tree. Every edge on $$P$$ is not a bridge of $$G$$, because they all lie on the cycle $$P\cup e$$. If there were more cycles in $$G$$, then there would be multiple paths from $$v$$ to $$w$$ in $$G-e$$, contradicting that it's a tree. Therefore $$P \cup e$$ contains all the non-bridges.