For that to be the case, in real hash functions, it would require you to be constrained to a certain number of possible preimages, not too many more than the nu

For that to be the case, in real hash functions, it would require you to be constrained to a certain number of possible preimages, not too many more than the number of possible hashes, so that it's likely that H^-1(H(x)) = x but there can still be a distinct y such that H(y) = H(x). So I concede that this could be the case under very specific circumstances, but I doubt it makes much difference in the real world.