# Nature of cycle space

In graph theory, the cycle space of a connected, undirected finite graph consists of all Eulerian (all vertices have only even degrees) subgraphs. It's easy to see why cycles generate (using symmetric difference as 'addition') only such subgraphs, but I cannot find a proof of the opposite, namely that each Eulerian subgraph is a member of the cycle space - how easy is it to show that? Since I have not received any answer yet, let me try my own proof: any Eulerian subgraph must contain at least one cycle (start at any positive-degree vertex of the subgraph and continue 'walking' through consecutively adjacent vertices till re-vising a vertex, and thus completing a cycle; this is always possible since every new vertex you enter has at least one extra incident edge to leave, until you re-visit a vertex). Then, remove all the edges of this cycle from the original subgraph; what remains is another Eulerian subgraph (since the degree of every vertex we have visited has been reduced by two). This can be repeated till no edges remain. This shows that a Eulerian subgraph consist of a union of cycles (potentially sharing vertices but not edges); such a union is clearly a member of the cycle space.