Exercise 7 solution:

(a) Suppose there exist integers p, q and a real x satisfying F^{q}(x) = x + p. Then f^{q}π(x) = πF^{q}(x) = π(x + p) = π(x), so that π(x) is a periodic point of period q for f.

(b) Suppose f^{q}(θ) = θ for some θ ∈ T. Let π(x) = θ. Then π(x) = f^{q}π(x) = πF^{q}(x), so F^{q}(x) = x + p for some integer p. Let π(y) = θ. Then y = x + k for some integer k, and F^{q}(y) = F^{q}(x + k) = F^{q}(x) + k = x + p + k = y + p. Hence there exists an integer p such that F^{q}(x) = x + p for all x ∈ π^{−1}(θ).