Then again, with your example, if you take time into account (otherwise it’s impossible), then EG2 is not from T1 to T0 but from T1(t=0) to T0(t=1), and your computational graph is acyclic. If fact, per definition, a computational graph is acyclic.
So, in your example, the graph looks like (no need to name edges) :
T0(t=0) ==> OP0(t=0) ==> T1(t=0) 
                           ||
                           ||
                           \/
                         T0(t=1) ==> OP0(t=1) ==> T1(t=1) 
                                                   ||
                                                   ||
                                                   \/
                                                T0(t=2) ==> OP0(t=2) ==> T1(t=2) 
                                                                         (...)

(as you may notice, I’m a painter in my free time)