Suppose we have $f:]a,b[ \rightarrow ]c,d[$ a $C^k$ function such that $f' > 0$. Then $f$ is a bijection and the inverse $f^{-1}$ is $C^k$.
¿How would I prove this as a consequence of the inverse function theorem? The inverse function obtains an inverse in the neighborhood of a point, but I don't know how to obtain an inverse for all the interval.
I think the way to go would be to prove that $f$ is a bijection (which I'm not sure how to do), and then taking into account that the inverse is unique, you can prove that at each point we can obtain a neighborhood where the inverse is locally $C^k$, but not sure how to argue that the patching would be $C^k$.
Becouse $\det f'(x)=f'(x)$ so we have $\det f'(x)\neq 0$ for all $x\in ]a, b[$ and thus $f$ satisfies the assumptions of inverse function theorem in each $x\in]a,b[$. So for all $x\in ]a, b[$ there is open set $V_x$ and $W_x$ such that $f$ is inverseable on $V_x$ and $f^{-1}:W_x\to V_x$ is $C^k$. Now, $$]a,b[ = \bigcup_{x\in]a,b[}V_x$$ and $$]c,d[ = \bigcup_{x\in]a,b[}W_x.$$ Of course if $g_1(y) = f^{-1}(y)$ for $y\in W_{x_1}$ and $g_2(y) = f^{-1}(y)$ for $y\in W_{x_2}$ then $g_1(y)=g_2(y)$ for $y\in W_{x_1}\cap W_{x_2}$.