Using some of the ideas in your construction I found a slightly smaller one on 8 nodes

I tweaked the example slightly as well but I don't think it was necessary.

This is a great find, but remember all the edge weights should be unique! Nonetheless it's interesting to know that the conjecture does not hold in general.

Edit: Also, is there both an edge (l0, b1) and (b1, l0) ? I may be misinterpreting

Btw if you're curious for context, this is related to a question in https://arxiv.org/abs/2108.00542 that Simple Stable Voting will return a candidate in the Split Cycle set over uniquely weighted tournaments.

f(c1)=c2 f(c2)=c3 f(c3)=l0
f(l0)=l1 f(l1)=l2 f(l2)=l0
f(b1)=l0 f(b2)=d1
f(d1)=l0

Edit: at the end of the row for l0, changed the last 3 values from 9/-7/- to -/9/-
Also changed the edge from l2 to b1 with weight -12 to an edge from b1 to l2 with weight -8.

The bottom graph shows the edges selected by f

This is a good observation. You can construct such a graph where f creates a a single cycle by, for example, having a Hamiltonian cycle consisting of all the strongest edges in E.

Unfortunately, f may not be injective, so it will not necessarily be a permutation. I believe, however, that we can reduce to the case where f(x) (for specifically the x as chosen to minimize (x, f(x)) has a unique preimage

Edit: If we take x such that (x,f(x)) is minimized, we can get a path from f(x) to x by repeatedly applying f and using the edge (f^i(x), f^i+1(x)). Since (x,f(x)) is minimal all of these edges will have weight higher than (x,f(x)). ((x, f(x)) will never have to be used because once x is reached, we're done.)
We are guaranteed to reach x because f is a permutation and permutations can always be decomposed into disjoint cycles.