I have the following problem: let $G$ be a finite directed graph with $V$ vertices $v_i$ and $E$ (directed) edges $e_j$. I know that if an edge $e_k = (v_i, v_j)$ is in the graph, then the opposite edge $-e_k = (v_j, v_i)$ is not in the graph. I also know that the graph contains at least one cycle. My goal is to render the graph acyclic by swapping the direction of some edges pertaining to at least one cycle.
I figured this was simple induction reasoning, i.e. swap an edge in any given cycle thus removing it, and repeat. But I cannot find a proof that it is possible to pick an edge which would not create a new cycle when swapped, nor can I find a counter-example with a finite graph.
The lemma I'd like to prove would be: given a graph $G$ such as above, there exists edge $(v_i, v_j)$ in one of the cycles so that replacing it with its opposite $(v_j, v_i)$ decreases the number of cycles in $G$ by at least 1. However, I couldn't find anything related... Any idea?
Thanks for your help