Condition for Edge to be Bridge

Theorem
Let $$G = \left({V, E}\right)$$ be a connected graph.

Let $$e \in E$$ be an edge of $$G$$.

Then $$e$$ is a bridge iff $$e$$ does not lie on any cycle of $$G$$.

Sufficient Condition
Let $$e$$ be a bridge of $$G$$.

Suppose, in the hope of generating a contradiction, that $$e = uv$$ and $$e$$ lies on a cycle, say $$C = \left({u, v, w, \ldots, x, u}\right)$$.

Thus $$G - e$$ contains a $u-v$ path, that is, $$\left({v, x, \ldots, w, u}\right)$$.

So $$u$$ is connected to $$v$$ in $$G - e$$.

Now we need to show that $$G - e$$ is connected.

Let $$u_1, v_1 \in V$$.

Since $$G$$ is connected, there is a $u_1-v_1$ path $$P$$ in $$G$$.

If $$e \notin P$$, then $$P$$ is also a path in $$G - e$$ and $$u_1$$ is connected to $$v_1$$ in $$G - e$$.

On the other hand, suppose $$e \notin P$$.

Then $$P$$ can be expressed as $$\left({u_1, \ldots, u, v, \ldots, v_1}\right)$$ or $$\left({u_1, \ldots, v, u, \ldots, v_1}\right)$$.

In the first case, $$u_1$$ is connected to $$u$$ and $$v$$ is connected to $$v_1$$ in $$G - e$$.

In the second case, $$u_1$$ is connected to $$v$$ and $$u$$ is connected to $$v_1$$ in $$G - e$$.

But we have already established that $$u$$ is connected to $$v$$ in $$G - e$$.

But Graph Connectedness is an Equivalence Relation, and hence $$u_1$$ is connected to $$v_1$$.

So, if $$e$$ lies on a cycle, then $$G - e$$ is connected.

This contradicts the stipulation that $$e$$ is a bridge.

Hence $$e$$ can not lie on a cycle.

Necessary Condition
Let $$e = uv$$ be an edge which does not lie on a cycle of $$G$$.

Suppose, in the hope of generating a contradiction, that $$e$$ is not a bridge.

Then $$G - e$$ is connected, and thus there is a $u-v$ path $$P$$ in $$G - e$$.

But $$P$$ together with $$e = uv$$ produces a cycle containing $$e$$.

This contradicts the stipulation that $$e$$ does not lie on a cycle of $$G$$.is a bridge.

Hence $$e$$ is a bridge.