Maximum number of edge-disjoint cycles vs minimum number of edges in a cut Let $G = \langle V,E \rangle$ be a directed graph, let $C(G)$ denote the maximal number of edge-disjoint cycles that can be packed into $G$, and let $D(G)$ denote the minimal size of a set $E' \subset E$ whose removal from $E$ results in an acyclic graph.
Clearly, $D(G) \geq C(G)$, because we must remove at least one edge from each cycle and the cycles counted by $C(G)$ are edge-disjoint. Is it true that $C(G) = D(G)$?
Seeing no reason why this should be true, I tried to construct a counterexample. Since I haven't been able to construct a counterexample, I suspect that the equality holds, but I have no idea how to prove it.

Alas! I found a counterexample with six vertices and nine edges.
Let $V = \{1,2,3,4,5,6\}$ and take $E = \{(1,2),(2,4),(4,3),(3,1),(3,5),(5,6),(6,4),(2,5),(6,1)\}$.
To see that this is a counterexample, first note that each vertex has total degree three, so no vertex can participate in more than one edge-disjoint cycle. Since the shortest cycle has length 4, this means that $C(G)=1$.
On the other hand, the three cycles $(1,2,4,3),(3,5,6,4),(1,2,5,6)$ don't have a common edge, and so you need at least to remove at least 2 edges to kill them off. Thus, $D(G) \geq 2$.
