Exercise 3 solution: Let n be an integer. Suppose there exist x_{1}, x_{2} ∈ [n, n + 1) with F (x_{1}) = F (x_{2}). Given F is a lift of f, πF (x_{1}) = πF (x_{2}) implies fπ(x_{1}) = fπ(x_{2}). Since f is injective, π(x_{1}) = π(x_{2}), implying x_{1} = x_{2} + k for some integer k. Given x_{1}, x_{2} ∈ [n, n + 1), k = 0, and we have x_{1} = x_{2}. Hence F is injective on [n, n + 1) for all integers n.

Suppose there exists an integer n such that F is strictly decreasing on [n, n + 1). Pick x_{1}, x_{2}, x_{3} ∈ [n, n + 1), x_{1} < x_{2} < x_{3}, and an integer m satisfying m < F (x_{3}) < F (x_{2}) < F (x_{1}) < m + 1. Note that, moving counterclockwise on T starting at π(x_{1}), we encounter π(x_{2}) before π(x_{3}). However, moving counterclockwise on T starting at πF (x_{1}) = fπ(x_{1}), we encounter πF (x_{3}) = fπ(x_{3}) before πF (x_{2}) = fπ(x_{2}). Now, since f is orientation preserving, starting at fπ(x_{1}) and moving counterclockwise we encounter fπ(x_{2}) before fπ(x_{3}), a contradiction. Thus F is strictly increasing on [n, n + 1) for all integers n.

We remark that F is strictly increasing on [n, n + 1] for all integers n. Let x ∈ [n, n + 1), and pick a sequence {x_{k}} ∈ [n, n + 1) with x_{k} < x_{k+1} for all k and lim_{k→∞} x_{k} = n + 1. Then F (x) < lim_{k→∞} F (x_{k}) = F (n + 1) by the continuity of F at n + 1. In particular, F (n + 1) > F (n), and so F (m) > F (n) for any integer m > n.

We now show F is strictly increasing on R. Let x_{1} < x_{2}, with x_{1} ∈ [n, n + 1) and x_{2} ∈ [m, m + 1) for some integers n ≤ m. If n = m then F (x_{1}) < F (x_{2}). If n < m then F (x_{1}) < F (n + 1) ≤ F (m) ≤ F (x_{2}), and we’re done.