I have some trouble prooving 1. to 2. and 2. to 3.
"Let $G$ be a connected graph. Prove that the next statements are equivalent:
- $G$ has a single cycle.
- $|E(G)| = |V(G)|$.
- $\exists \ e \in E(G)\ $ such that $\ G - e\ $ is a tree."
I already have a $\textbf{Lema}$ that says "Let $G$ be a connected graph and $C$ a cycle in $G$. if $e$ is an edge of $G$, then $G - e$ is connected" and it maybe can be used to proove 1. to 3. , but I don't know how to continue with the other proofs.
$\bf1\to3$: If $G$ has a single cycle, we can break it by removing any edge $e$ in the cycle without disturbing the connectedness of $G$. $G−e$ is a connected, acyclic graph i.e. a tree.
$\bf3\to2$: Given for some $e$, $G-e$ is a tree. The number of edges in a tree is one less than the number of vertices. $|E(G-e)|=|E(G)-\{e\}|=|E(G)|-1$. Note that we don't remove any vertex when we remove an edge, so $|V(G-e)|=|V(G)|$. thus $|V(G)|-1=|E(G)|-1$, giving us $|V(G)|=|E(G)|$.
$\bf2\to1$: If $G$ is acyclic, then $G$ is a tree. So $|E(G)|=|V(G)|−1$, which is a contradiction. So $G$ has at-least one cycle. Let $C_1,C_2$ be cycles in $G$. We can remove an edge $e_1\in E(C_1)$ without disturbing the connectedness of $G$. $G-e_1$ is connected and $|E(G-e_1)|=|E(G)|-1=|V(G-e_1)|-1$ so $G-e_1$ is a tree. Since trees are acyclic, $e_1\in C_2$. Since $e_1$ is arbitrary, we conclude $C_1=C_2$ and $G$ has a single cycle.