For a $2$-edge-connected simple graph $G$ and a tree $T$ of $G$, let $C_e$ be the unique cycle in $T + e$, $e \in E(G) - E(T)$. Define the set $\mathcal{C}(T) = \{C_e | e \in E(G) - E(T)\}$.
Now given a set of cycles $\mathcal{C}$ of $G$, is it possible to decide if $\mathcal{C}$ = $\mathcal{C}(T)$ for some tree $T$ of $G$?
EDIT:
It is easy to see that the following two conditions for $\mathcal{C}$ = $\mathcal{C}(T)$ are necessary :
$|\mathcal{C}|$ = $|E(G)| - |V(G)| + 1|$
The set $\{e \in E(G)|\exists!c\in \mathcal{C}, e \in c \}$ is forest of $G$
$|\mathcal{C}|$ = $|E(G)| - |V(G)| + 1|$
1') $\cup\mathcal{C} = E(G)$
- The set $\{e \in E(G)| e \in c_1\cap c_2 \text{ and } c_1,c_2 \in \mathcal{C} \text{ and }c_1 \ne c_2 \}$ is forest of $G$
My question is: Are they also sufficient?