Just need help proving the statement below, for which I've outline how the proof would go:
Theorem: An edge $e$ of a graph $G$ is a bridge iff $e$ lies on no cycle of $G$
Attempt $(\implies)$
Assume for the sake of contradiction that there exists a bridge $e$ in the edge set of $G$, such that $e$ lies on a cycle. Then, if we take a simple case such as $C_{3}$, we arrive at $P_{3}$, which is connected. This is a contradiction, and so there cannot exist an edge $e$ such that $e$ is a bridge that lies on a cycle.
Attempt $(\impliedby)$
Let $e$ be any edge such that $e$ lies on a cycle, $C_{n}$. Then, let us remove $e$ from the edge set of $C_{n}$. By definition of a cycle (a closed path) we know that the removal of any single edge in the cycle creates a path $P_{n}$. A path is connected, and so we arrive at the conclusion that the removal of any single edge is not enough to disconnect a cycle.
Any feedback would be great!
Best Answer
You need to be a bit more precise IMHO. Introduce notation.. Let $e$ be an edge that is a bridge and suppose it lies on a cycle $C$. So let $C$ consist of points $x_0,\ldots,x_n, x_{n+1} = x_0$ such that $(x_i, x_{i+1})$ are edges of $G$ for all $i =0, \ldots n$. We can suppose WLOG (renumbering if needed) that $e$ is the edge $(x_n, x_0)$. If we remove $e$ from the graph, the number of connected components must increase, and the only component $G'$ of $G$ that could get split is the one containing $e$, but it's easy to see it does not get split, but stays connected: if there was a path between two points $p,q$ in $G'$ that contained $e$, then we still have a path without it, just take the route $x_0 \to \ldots \to x_n \to x_n$ instead of $e$. So $e$ is not a bridge: contradiction.
Essentially the same argument proves the converse as well: a single edge on a cycle can always be replaced by the rest of the cycle.