A graph G containing cycle C, how to prove G' = G - e?

Question

If there is a connected graph $G$ that containing a cycle $C$ and let $e = (u, v)$ be any edge that lies on this cycle. How can we prove that graph $G' = G - e$ if the sub-graph obtained from $G$ by removing the edge $e$ is also connected?

Answer 1

Here is a hint.

A graph is said to be connected if for all distinct pairs of vertices $u$ and $v$ in it there exists a path from $u$ to $v$. To show $G'$ is connected consider the path $P$ in $G$ which connected $u$ to $v$, and see if we can reuse them.

• If $P$ doesn't contain the edge $e$ then you can use it to get to $v$ from $u$.
• If $P$ does contain the edge $e$, you can create a new path $P'$ by ....

The meat of the argument is showing you can work around not having the edge $e$.

If you get stuck consider some simple graphs $G$ such a cycle.

Answer 2

I shall only prove part of it for you. The rest is to be done by you(the OP).

Suppose the graph $G$ consists of $n$ vertices in $V = (v_1,v_2...,v_n)$ and the corresponding edge set is $E = (e_1,e_2....e_m)$. Let $c \subset V$ denote the vertices which form the cycle in $G$, and $p \subset E$ denote the edges in the cycle. Clearly, a cycle is connected(try to prove this yourself). Represent a graph as: $c = c_1 - c_2-.....-c_k$ [where $c_1-c_2$ denotes that there exists an edge connecting these 2 vertices]. For a cycle, $c_1 = c_k$, and all other vertices $c_i ; i \ne 1$ occurs exactly once. Clearly, we can see that there exists 2 distinct paths between $c_1$ and $c_j$ for any $j \ne 1$ [in the representation, the 2 paths are the subsequences formed by starting from either end and reaching upto vertex $c_j$.

Also, note that if $v_i \in c$, then there exists at least 2 distinct vertices $v_k,v_l \in c$, which are connected to $v_i$. Now, Suppose that some $e_i \in p$, connecting $v_k$ and $v_l$ is removed. We must prove that the graph $c - e_i$ is still connected. Label $v_k = c_1$ and $v_l = c_r$. As per the proof in the last paragraph, there exists 2 paths $P_1$ and $P_2$ between $c_1$ and $c_r$. Suppose $e_i \in P_1$. $=> e_i \notin P_2$ (can you justify why?) Thus, $c_1$ and $c_r$ are still connected and so are the other vertices. Thus $c-e_i$ is connected. Now,as an exercise prove that any graph $G$ containing a cycle $c$ is also connected, on removing some $e_i \in c$.