Exercise 4 solution: We use induction to show that πF^{n}(x) = f^{n}π(x) for all x ∈ R and all positive integers n . Let x ∈ R. The n = 0 case is π(x) = π(x). Suppose πF^{n}(x) = f^{n}π(x) for some n > 0. Then πF^{n+1}(x) = πF (F^{n}(x)) = fπ(F^{n}(x)) = f(f^{n}(π(x)) = f^{n+1}π(x). Thus F^{n} is a lift of f^{n} for all positive integers n.

By Exercise 3, F^{−1} : R → R exists. Moreover, πF = fπ implies f^{−1}π = πF^{−1}, so that F^{−1} is the lift of f^{−1}. Let n > 0. Since F^{−n} is the nth iterate of F^{−1}, F^{−n} is a lift of f^{−n} by the argument above. Hence F^{n} is a lift of f^{n} for all integers n.