real analysis - Properties of a one to one continuous function from $[0, 1]$ onto itself

Let $f$ be a one to one continuous function from $[0, 1]$ onto itself. Show that

(i) $f$ is a homeomorphism.

(ii) $f$ is strictly monotone on $[0, 1]$

(iii) Is it true that if $f$ is strictly monotone on $[0, 1]$ and onto $[0, 1]$, then $f$ is continuous?

For (iii) I am sure it will be true as monotone function has only jump discontinuity so if it is onto it has to be continuous as well. This is my intuition only, more rigorous argument is welcome, thank you.

Answer

i) Because $[0,1]$ is compact and Hausdorff, $f$ is a homeomophism.

ii) Any one-one continuous function (on $\Bbb R$ to $\Bbb R$) is strictly monotone.

iii) If $f$ is not continuous, at some point $a\in (0,1)$ we have:
$$\lim_{x\to a^-}f(x)\ne \lim_{x\to a^+}f(x)$$
$$\Rightarrow \sup_{x$$\Rightarrow \sup f([0,a))< \inf f((a,1])$$
So $f([0,1])\ne [0,1]$ .