I was asking to prove that
For any directed graph, when we perform BFS, for each cross edge $(u,v)$, there is some time when both vertices $u$ and $v$ appear in the queue. (In other words, we have a queue like this: $Q = \langle v_1,v_2,..u,..,v,..v_k \rangle$.)
However, when I draw some graph; I've encountered this example
If we perform BFS starts from $a$, then what is the edge of $(f,c)$? clearly, if what I was about to prove is correct, $f$ and $c$ will never appear in the queue at same time during the whole execution of BFS. So $(f,c)$ is not cross edge. Then what edge is this ? It can't be back edge; since $c.parent$ is $a$, and $f.parent$ is $e$. they aren't tree edge, nor forward edge.
So what I am asked to prove is not correct?
So I guess that the statement only holds for undirected graph?