# Is this inverse function continuity proof consistent?

I am self-studying Calculus and tried to solve the following question:

## Homework Statement

Suppose that the function f is continuous and increasing in the closed interval [a, b]. Then
(i) f has an inverse f-1, which is defined in [f(a), f(b)];
(ii) f-1 is increasing in [f(a), f(b)];
(iii) f-1 is continuous in [f(a), f(b)].
The book does the proofs for (i) and (ii). It then proofs (iii) partially by showing that f is continuous in the open interval (a,b).
I must then show that f-1 is continuous in f(a) to the right and continuous in f(b) to the left.

## Homework Equations

Definition of unilateral limit to the right:
$$\lim_{x\to a^+} f(x) = L$$
if and only if, for every small ε > 0, there is a δ > 0 such that
|f(x) - L| < ε whenever 0 < x - a < δ

## The Attempt at a Solution

I will try to do here the proof that f-1 is continuous in f(a) to the right.
Applying the definition of unilateral limit to the right, I must then show that, for every small ε > 0, there is a δ > 0 such that
|f-1(y) - a| < ε whenever 0 < y - f(a) < δ

I begin by defining ε such that a + ε belongs to [f(a), f(b)].
Since a < a + ε, then, since f is increasing:
$$f(a) < f(a+\epsilon)$$
Now, since the choice of δ depends upon the choice of ε, I choose δ such that:
$$\delta \leq f(a+\epsilon) - f(a)$$
$$f(a)+\delta \leq f(a+\epsilon)$$
Now, from the definition of the limit above, I know that:
$$0 < y - f(a) < \delta$$
Summing f(a) to every term:
$$f(a) < y < f(a) + \delta$$
So, combining this with the δ chosen above:
$$f(a) < y < f(a) + \delta \leq f(a+\epsilon)$$
Since f and f-1 are both increasing:
$$f^{-1}(f(a)) < f^{-1}(y) < f^{-1}(f(a+\epsilon))$$
$$a < f^{-1}(y) < a+\epsilon$$
Subtracting a from every term:
$0 < f^{-1}(y) - a < \epsilon$ whenever $0 < y - f(a) < \delta$
which appears to be the desired result, except for the lack of absolute value. I imagine the absolute value didn't appear here because $f^{-1}$ is an increasing function.

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