Can someone please help. I am trying to prove the following.
Prove that an edge {u,v} in a connected graph G is a bridge if and only if every chain from u to v in G includes edge {u,v}.
in this text chain is the same as a path. I am at the following with my proof.
--Suppose that edge {u,v} is a bridge in a connected graph G and there exists a chain that does not contain edge {u,v}. Then since the chain must be connected by definition, but the chain is a subgraph of G \ {u,v} is a disconnected graph thus the chain contains a bridge.
Am I on the right track? Any thoughts would be helpful.
Thanks,
The first thing to observe is that if $\{u,v\}$ is a bridge then there is no $uv$-path in $G\setminus uv$.
You have started your argument correctly (sentence 1). After that you should observe that $G\setminus uv$ contains the $uv$-path that doest not use $uv$. What can you conclude now?
Proving the converse should be easier.