Exercise 15 solution: We are given hf(θ) = gh(θ) for all θ ∈ T, where h is a homeomorphism.

(a) We prove by induction: hf^{n}(θ) = g^{n}h(θ) for all θ ∈ T and all n ≥ 0. Let θ ∈ T. The n = 0 case yields h(θ) = h(θ). Suppose hf^{n}(θ) = g^{n}h(θ) for some n > 0. Then hf^{n+1}(θ) = hf^{n}(f(θ)) = g^{n}h(f(θ)) = g^{n}gh(θ) = g^{n+1}h(θ), and we are done.

Note that hf = gh implies g^{−1}h = hf^{−1}, so that f^{−1} and g^{−1} are also conjugate. Let n > 0. Since f^{−n} is the nth iterate of f^{−1}, we have hf^{−n}(θ) = g^{−n}h(θ) by the argument above. Thus hf^{n}(θ) = g^{n}h(θ) for all n ∈ Z.

Let f^{n}(θ) ∈ o(θ, f). Then hf^{n}(θ) = g^{n}h(θ) ∈ o(h(θ), g), so h(o(θ, f)) ⊂ o(h(θ), g). Let g^{n}h(θ) ∈ o(h(θ), g). Then g^{n}h(θ) = hf^{n}(θ) ∈ h(o(θ, f)), so o(h(θ), g) ⊂ h(o(θ, f)). Hence h(o(θ, f)) = o(h(θ), g).

(b) Suppose o(θ, f) is dense in T. Let I be an open interval (arc) in T. Since h is a homeomorphism, h^{−1}(I) is an open set in T. Pick n ∈ Z with f^{n}(θ) ∈ h^{−1}(I). Then hf^{n}(θ) = g^{n}h(θ) ∈ I, and o(h(θ), g) is dense in T.

For the other direction, note that g is conjugate to f via the homeomorphism h^{−1}, so o(θ, g) dense in T implies o(h^{−1}(θ), f) is dense in T.