The differentiability class of the inverse function Here's the final part of a proof (from Marden's Elementary Classical Analysis) of the inverse function theorem, where we have been given that $f$ is of class $C^p$:

Could someone please explain the argument starting from the third line from the bottom ("$Df$ is of class $C^{p-1}$...")?
 A: Let's put the argument on a more abstract footing. You have a map
$$F = G \circ H \circ I.$$
You know that $G$ is $C^\infty$, $H$ is $C^k$, and $I$ is $C^m$. Then you immediately know that $F$ is (at least) of the class $C^{\min \{k,m\}}$.
Here, we have $F = D(f^{-1})$, $G$ the inversion of a linear map, $H = Df$, and $I = f^{-1}$, and we start knowing that $k = p-1$, and $m = 0$.
That yields that $F\in C^0$. But $F$ is the derivative of $I$, hence in fact $I \in C^1$. Now, if $k \geqslant 1$, the abstract argument shows that $F$ itself is of class $C^1$. But since $F$ is the derivative of $I$, that means $I \in C^2$. If $k \geqslant 2$, we then have $F\in C^2$, and that implies $I \in C^3$. So while the known regularity ($C^m$) of $I = f^{-1}$ is less than $C^p$, from the representation of $D(f^{-1})$ as a composition, it follows that $D(f^{-1})$ itself is $C^m$-regular, which implies that $I = f^{-1}$ is $C^{m+1}$-regular. The chain stops when the regularity of $H = Df$ is exceeded, since then $\min \{p-1,m\} = p-1$ and knowing higher regularity of $G$ and $I$ would not guarantee higher regularity of $F$ anymore.
