I believe that I have a result that is well known by someone here. If you know where I can find a proof, then I would appreciate it. It seems like elementary graph theory, but I have not been able to find it or prove it myself.

Here is the set-up. Let $n$ be a positive integer and let $G$ be a 2-connected, cubic (3-regular), undirected multigraph with vertex set $\{1,2,3,...,2n\}$. Thus, $G$ has precisely $3n$ edges, no bridges, but it might have a cut set with only 2 edges. Next, suppose $T$ is a spanning tree of $G$. Then the complement $G\backslash T=\{e_0,e_1,e_2...,e_n\}$ consists of precisely $n+1$ edges. For each edge $e_i\in G\backslash T$, there is a unique cycle, say $\gamma_i$, in $T\cup e$. Define a function $f$ from the edge set $E(G)$ to the power set $\mathcal{P}(\{0,1,2,...,n\})$ as $$f(e)=\{i\in\{0,1,2,...,n\}: e \in \gamma_i\}.$$ Thus, $f$ tells which which cycles among $\{\gamma_0,\gamma_1,\gamma_2,...,\gamma_n\}$ pass through a given edge.

With this set-up, here is my observation: Such a graph appears to be 3-connected if and only if $f$ is one-to-one. That is, given a graph $G$, one can detect a 2-edge cut set exactly when there are two edges $e_1$ and $e_2$ that have the same cycle assignments. The choice of the spanning tree used to define $f$ seems to be irrelevant as far as this goes. That is, the function $f_T$ always seems to detect the same 2-edge cut sets for any choice of $T$.

(I should add that this observation seems to hold for multigraphs that are not 3-regular, if only 2-connected. For my purposes, however, I only care about 3-regular graphs.)

Do you know how to prove or disprove this observation? If you can point me to a reference, then I would appreciate it.