Why can I remove an edge from a cycle that is part of a connected graph I've been trying to figure this out for awhile now. But if I have a connected graph, which contains a cycle, why can I remove an edge from the cycle and still have a connected graph?
 A: Partition the vertices that form a cycle into two disjoint sets, $L$ and $R$ such that all nodes in $L,R$ respectively form sub-paths that are part of the cycle. Let the other nodes that are not part of the cycle be in $N$.

There should be exactly two edges that connect the nodes from $L$ to $R$ since the set of node in $L \cup R$ form a cycle. Removing one edge still leaves $L$ connected to $R$ which means the graph is still connected because the remaining nodes in $N$ are connected to either $L$ or $R$.
A: Let $G$ be a connected graph and suppose $C$ is a cycle of $G$.

Consider the subgraph $G-e$, where $e=uv$ is an edge of $C$.

Removing $e$ converts $C$ into a path from $u$ to $v$ having the same vertices as $C$,  hence the vertices of $C$ are all in the same component, $A$ say, of $G-e$. 

But if we then add edge $e$ back, it will be an edge connecting the two vertices $u,v$ of $A$, so adding $e$ won't change the number of components. 

Thus in the transition from $G-e$ back to $G$, the number of components remains the same, hence since $G$ is connected, so is $G-e$.
