1. question on a proof

I need a little confirmation on the following proof:
The problem is this:
For an interval I, with f:I -> R where f is strictly increasing with f(I) an interval, prove that f^(-1):R -> I is strictly increasing.

My proof is this:
Because f is strictly increasing, for y,z exists in I such that y<z,
f(y)<f(z).
Let y,z exist in I. Then f(y),f(z) exist in R. and f(y) < f(z).
but f^(-1)[f(y)]<f^(-1)[f(z)] which implies y<z and thus f^(-1) is strictly increasing.
This proof seems trivial to me, which makes me think I did something wrong. I possibly assumed too much in my proof, not sure. Any ideas?

2. Hello

I suffer from similar habbits. Always assuming things are complicated. Your proof is correct in its simplicity. You are absolutely right to assume the existence of $f^{-1}$, its given in the problem statement. Hence you may use the identity function $f^{-1}(f(x)) = x$, by the definition of the inverse.

In fact just to reassure you that your proof makes sense, remember that f is the inverse of the function $f^{-1}$. And a function only has an inverse if it is strictly increasing or decreasing (a standard theorem from set theory). You have of course proved $f^{-1}$ is strictly increasing. And thus the inverse of $f^{-1}$ exists. namely, f.