Fundamental Cycles of a graphs

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 conditions for $\mathcal{C}$ = $\mathcal{C}(T)$ are necessary :

1) $|\mathcal{C}|$ = $|E(G)| - |V(G)| + 1|$

1') $\cup\mathcal{C} = E(G)$

2) 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?

• This looks easy enough to be a homework question. Any motivation for why you're asking? – David Eppstein Nov 28 '13 at 5:02
• It is not a homework, but I should have been more precise(see edit.) I am interested in studying the correspondence between tree and even graphs arising from the symmetric difference of their Fundamental cycles. – hbm Nov 28 '13 at 20:55
• Before (2) you should add that every edge is in at least one of the cycles. Another necessary condition is that the intersection of any two cycles is empty or a path; does it follow? – Brendan McKay Nov 28 '13 at 21:36

• Yes, a largest set, and they're all trees iff they all have exactly $n-1$ edges. – David Eppstein Dec 4 '13 at 7:30