Identifying the edges that are essential for biconnectivity Question:
If $G(V,E)$ is a biconnected symmetric graph, is it possible to identify the edges, whose deletion destroys biconnectivity, in the following way:  


*

*determine the union $B:= ST\cup F_1$ of a spanning tree and a maximal forest, that is edge-disjoint with $ST$

*determine a second maximal forest $F_2$ that has no edges in common with $B$. 

*take as the critical edges those edges of $B$, that are adjacent to a vertex of degree 2 or connect different trees of $F_2$.

 A: This seems like a counterexample: take 3 cliques $C_1$, $C_2$ and $C_3$, each on say $5$ vertices. Connect $C_1$ and $C_2$ with three vertex-disjoint edges, and connect $C_2$ and $C_3$ with 2 edges. Now it is possible to choose $ST$ to be a spanning tree going from $C_3$ to $C_2$ to $C_1$ and back to $C_2$, and $F_1$ a disjoint spanning tree going from $C_1$ to $C_2$ to $C_3$, so that the following holds: the remaining graph has three connected components, $F_2$ has a tree in every component, and $B$ connects the trees in $C_1$ and $C_2$. Hence the procedure will return an edge between $C_1$ and $C_2$, which is not critical.
A: Given a bi-connected graph $G$, say an edge $e$ is destructive if $G-e$ has a cut-vertex. Say that $e$ is  $(T,F_1,F_2)$-critical if there is a spanning tree $T$ and forests $F_1,F_2$ as in you description with respect to which $e$ critical.


*

*Must a $(T,F_1,F_2)$-critical edge be destructive? The answer is no. Here is a example based on the answer given by  @smapers.


Here is a $14$ point graph with $38$ edges. If desired, add another $4$ edges on the left and $3$ on the right to make each a $K_7$. Either way, there are no destructive edges.  The tree $T$ is the green path, The forest $F_1$ is the red tree And $F_2$ consists of two blue paths. Then the central red edge is critical but not destructive.

If one wants $G$ to have some destructive edges and be exactly bi-connected then, as suggested, add some new vertices and edges making a $K_5$, connect it to the $K_7$ on the left two new edges which will become part of $T$ (which will lose an edge elsewhere to avoid a cycle) and then add enough edges to make $F_1$ and $F_2$ valid.
This illustrates:


*Do the $(T,F_1,F_2)$-critical edges include all the destructive edges? Again, no. One can certainly arrange to have some of the destructive edges in $T.$

*If $e$ is destructive is there some way to choose the tree and forests so that it is critical? Yes. Assign $e$ to $F_1$ with the rest to be determined. Now $G-e$ is connected with a cut-vertex $v$. Start with all the edges on $v$ and extend that to a spanning tree $T$.  Now $G-e-T$ is disconnected with the vertices of $e$ in distinct connected components. So $F_2$ will have those two vertices in distinct distinct trees. How ever we form $F_1$ (using $e$ ) and $F_2$, The destructive edge $e$ is critical.
Note that some of the trees in $F_2$ might be isolated vertices.
