how to show strictly increasing function on an interval has continuous inverse my question is to show that a strictly increasing function that is defined on an interval has a continuous inverse.
Thanks.
 A: There is not necessary to have the continuous condition for $f(x)$. Consider the example given by Betty Mock, $f^{-1}(y)$ is not continuous in $\mathbb{R}$ but it is continuous in the domain of $f^{-1}$, i.e. $[0,1/2]\cup (3/2,2]$. 
In the discontinuous domain, the continuity of function become less intuitive because by the definition of function is,$\forall \epsilon$, $\exists \delta$,s.t. $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$, where $x$ and $a$ are both form domain $D$. 
A: You have to show that $f:I\to f[I]$, where $I$ is the interval, is an open map. It suffices to show that each interval of the form $(a,b)\ a,b\in I$ has an open image. Try to show that 
$f[(a,b)]=(f(a),f(b))$. If $I$ is of the form $[i,s)$, then you also need to show that $f[[i,b)]=[f(i),f(b))$. Similarly, if $I$ has a maximum. This is because those intervals form a basis for the open subsets in $I$.
A: If it is not continuous neither is its inverse. However, every strictly increasing function will have an inverse on the domain where it is strictly increasing.  
The requirement of an inverse is that f($f^{-1}$(x)) = x.  Since f is strictly increasing it is a 1 to 1 mapping onto its range.  The inverse is simply the function that maps each element of the range back into the element that generated it in the domain.  That is, if f(x) = y then $f^{-1}(y)$ = x.
If the function f is continuous, you can prove its inverse is continuous.  Working from the basic definition of continuous:
Let a be a point in the interval and suppose that for every sequence $x_n \rightarrow a$ we have $f(x) \rightarrow f(a)$, so that f is continuous at a.  We want to show that $f^{-1}$ is continuous at f(a).  However by definition $f^{-1}(x_n) = x_n \text{ and } f^{-1}(a) = a$.  So $f^{-1}(f(x_n)) = x_n \rightarrow a = f^{-1}(f(a))$ and $f^{-1}(y)$ is continuous at y = f(a).
A: First suppose $I:=]a,b[$. Since $f:I\to\mathbb{R}$ is strictly increasing, it is injective, so $f:I\to f(I)$ is a bijection and $f^{-1}$ exists. $f^{-1}$ is continuous iff $f$ is an open map. It suffices to check that $\forall ]\alpha,\beta[\subseteq I: f(]\alpha,\beta[)$ is open in $f(I)$.
Let $]\alpha,\beta[\subseteq I$. We claim that $f(]\alpha,\beta[)=]f(\alpha),f(\beta)[\cap f(I)$. If $\alpha<x<\beta$, then $f(\alpha)< f(x) < f(\beta)$ and since $x\in ]\alpha,\beta[\subseteq I, f(x)\in f(I)$. Conversely, if $y\in ]f(\alpha),f(\beta)[\cap f(I)$, there is an $x\in I: f(x)=y$. Also since $f(\alpha)<f(x)=y<f(\beta), x\in ]\alpha,\beta[$, so that $y\in f(]\alpha,\beta[)$. This means that $f(]\alpha,\beta[)$ is an open subset with respect to the subspace topology of $f(I)\subseteq\mathbb{R}$. Therefore $f$ is open.
Other cases follow analogously.

P.S.: For reference purposes, this is Exercise 2.7.41 from Royden & Fitzpatrick's Real Analysis, 4e (p. 53).
