Question 6.T.8: Let f : I → R, where I is an interval. If f is continuous an...

By Theorem 6.7, f is strictly monotonic. To see that f^{−1} is also monotonic, suppose f is strictly increasing and y_{1}, y_{2} ∈ f(I) with y_{1} < y_{2}. If x_{1} = f^{−1}(y_{1}) and x_{2} = f^{−1}(y_{2}), then x_{1} < x_{2}; for the inequality x_{1} ≥ x_{2} implies

y_{1} = f(x_{1}) ≥ f (x_{2}) = y_{2},

since f is increasing. The continuity of f^{−1} : f (I) → I follows from Lemma 6.1, I and f(I) being intervals.