I got this question, and I'd be happy for help.
G=(V,E). $G_S$ is a Strongly-Connected Components's Graph.
I need to prove, that if there is only one Component ($C_0$) which is not with incoming edge, that there is a DFS running in G that ruturn one tree, and only him.
I thought to start from the vertex with the lattest finish time in $C_0$. It works on my examples, but I dont know how to prove it, and if it is correct.
Thank you for your help.