strictly increasing and continuous and onto function.

Question

Why if we have strictly increasing, continuous and onto function its inverse must be continuous? could anyone explain this for me please?

Answer 1

When a function is one-one and onto, or in other words the function is bijective, there exists an inverse for the function. Just because the inverse of the function is the mirror image of the function itself about the line $y=x$, we must expect the same behaviour as of the original function, but in an inverted sense.

So if the original function is continuous, by basic logic, its mirror image must also be continuous.

Answer 2

If $f$ is strictly increasing and onto, it must be continuous.

To see this, suppose we pick some $x_0$ and $\epsilon>0$. Since $f$ is onto, there $a,b$ such that $a<x_0<b$ and $f(a) = f(x_0)-{1 \over 2} \epsilon$ and $f(b) = f(x_0)+{1 \over 2} \epsilon$. Choose $\delta = \min (|a-x_0|,|b-x_0|)$, then if $x \in (x_0-\delta,x_0+\delta)$ we have $|f(x)-f(x_0)| < \epsilon$.

If $f$ is strictly increasing and onto, then it must have an inverse and the inverse is also strictly increasing and onto, hence continuous by the above.