Exercise 5 solution: Suppose πF (x) = fπ(x) and πG(x) = fπ(x) for all x ∈ R. Then given a real x, there is an integer k_{x} such that F (x) = G(x) + k_{x}. The function d : R → Z, d(x) = F (x) − G(x) is continuous since F and G are both lifts of f. Since R is connected, d(R) ⊂ Z is connected. The only connected subsets of Z are singleton sets; therefore d(R) = {k} for some integer k. That is, there is an integer k satisfying F (x) = G(x) + k for all x ∈ R.