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?



Fundamental Cycles of a graphs
I had a complicated algorithm based on matroid intersection here earlier, but there's a much simpler answer: a cycle basis is fundamental if and only if each cycle includes at least one edge that is not part of any other cycle. And it's easy to see that any collection of cycles with this property is automatically linearly independent. So you can check if a collection of cycles is a fundamental cycle basis by verifying that each has a unique edge and that there are enough cycles to form a basis.


The reference is:


Cribb, D. W.; Ringeisen, R. D.; Shier, D. R.
On cycle bases of a graph. 
Proceedings of the Twelfth Southeastern Conference on Combinatorics, Graph Theory and Computing, Vol. I (Baton Rouge, La., 1981). 
Congr. Numer. 32 (1981), 221–229. 


