Exercise 6 solution:

(a) Let x ∈ R. Then πF (x+1) = fπ(x+1) = fπ(x) = πF (x), so F (x+1) = F (x)+k for some integer k. Since F is strictly increasing, k ≥ 1. Suppose k > 1, so that [F (x), F (x) + 2] ⊂ F ([x, x + 1]). Let y ∈ (F (x), F (x) + 1). By the Intermediate Value Theorem and the fact F is strictly increasing, there exist x_{1}, x_{2} ∈ (x, x + 1), x_{1} < x_{2}, with F (x_{1}) = y and F (x_{2}) = y +1. Note that π(x_{1}) ≠ π(x_{2}). We have πF (x_{1}) = π(y) and πF (x_{2}) = π(y + 1) = π(y), implying fπ(x_{1}) = π(y) = fπ(x_{2}), contradicting the fact f is injective. Hence k = 1, and F (x + 1) = x + 1 for all reals x.

(b) We show by induction that for all reals x and for all positive integers k, F (x+k) = F (x) + k. This is clearly true when k = 0. Suppose F (x + k) = F (x) + k for some k > 0. Then F (x + k + 1) = F (x + k) + 1 = F (x) + k + 1, and we’re done.

Now let k < 0 be an integer. Since −k > 0, we have F (x) = F (x + k − k) = F (x + k) − k, or F (x + k) = F (x) + k. Thus F (x + k) = F (x) + k for all integers k.

Let n be an integer, and set H(x) = F^{n}(x) and h(θ) = f^{n}(θ). Let k be an integer. Since H is a lift of h by Exercise 4, F^{n}(x + k) = H(x + k) = H(x) + k = F^{n}(x) + k.