Let $G$ be a $2$-connected graph and for $e \in E(G)$ denote by $\mathcal{C\_e}$ the set of all cycles of $G$ containing the edge $e$.


For what set of edges does $\mathcal{C\_e}$ contain a basis of the cycle space of $G$?



Do the cycles containing a fixed edge generate the cycle space of a graph?
For all $e \in E(G)$, $\mathcal{C}\_e$ contains a basis of the cycle space. 


**Proof.** Let $e=uv \in E(G)$, and let $C$ be a cycle of $G$. If $e \in E(C)$, then there is nothing to show. Thus, $e \notin E(C)$. By Menger's theorem, there are two disjoint paths $P\_u$ and $P\_v$ from $\{u,v\}$ to $V(C)$. By taking $P\_u$ and $P\_v$ to so that $|P\_u|+|P\_v|$ is minimal, we may assume that $P\_u$ and $P\_v$ each meet $C$ once. Thus, $e \cup E(P\_u) \cup E(P\_v) \cup E(C)$ is a $\theta$-subgraph, and it is easy to see that $C$ is the symmetric difference of two cycles $C\_1$ and $C\_2$, both of which contain $e$. 


