# Shortcut in proof of continuity/differentiability in inverse function theorem

The messiest, least interesting part of the various proofs of the inverse function theorem comes after you have constructed the inverse function and must now establish continuity and differentiability.

The estimates required are not very hard, but it has always bothered me aesthetically.

While pondering this today, I realized that these estimates could be bypassed if continuity and differentiability were defined in terms of topological and geometrical properties of the graph. If $f : X \subset \mathbb{R}^m \to \mathbb{R}^n$ is invertible then $\Gamma_f = \{(x, f(x)) : x \in X\} \subset \mathbb{R}^m \times \mathbb{R}^n$ and $\Gamma_{f^{-1}}$ are related by an isometry, so a characterization of continuity and differentiability in terms of a function's graph would immediately imply that $f$ is continuous or differentiable iff the same is true for $f^{-1}$.

With a single variable, we have the characterization that $f : \mathbb{R} \to \mathbb{R}$ is continuous iff its graph is connected and closed in $\mathbb{R}^2$. For differentiability, we can consider the pencil of lines through a point $x$ and say that $f$ is differentiable at $x$ iff the graph is pathwise connected in a neighborhood $U \subset \Gamma_f$ of $x$ and if for all sequences $\{x_i\}$ in $U$ converging to $x$ the corresponding secants all converge (in the projective space structure) to the same limit.

With this, we have a completely qualitative proof of the inverse function theorem for one variable. Assume $f$ is continuously differentiable and that $f'(x_0) \neq 0$. By continuity of $f'$, we see that $f'$ has constant sign in a neighborhood of $x_0$. Therefore $f$ is locally monotone and hence invertible on that neighborhood. As f is continuous and differentiable and we've defined these properties in terms of graphs, it follows that this local inverse is likewise continuous and differentiable since the two graphs are reflections.

What about definitions for the general case of $f : X \subset \mathbb{R}^m \to \mathbb{R}^n$?

Well, the definition of differentiability is easily extended. Instead of sequences of single points, we must use sequences of $m$-tuples of points and require that the points in each tuple are affinely independent so they span an $m$-flat in $\mathbb{R}^m \times \mathbb{R}^n$. The pencil of lines is replaced with the pencil of $m$-flats carrying the standard structure as a Grassmannian. Straightforward stuff, I think.

The thing I'm unsure about is whether the graph-based definition of continuity can be extended to several variables. The definition using closedness and connectedness works for $m = 1$ since $\mathbb{R}^n$ is locally compact, but for larger $m$ there are counterexamples already with $m = 2$ and $n = 1$.

1. Can the definition of graph-based continuity be repaired to work for all $m$ and $n$? At least in the $n \geq m$ case we care about for the IFT? Maybe the one based on closedness and connectedness is actually adequate if we presuppose graph-based differentiability as well?
2. Do you see any holes in my reasoning?
• It appears that $f(x)=x^{1/3}$ is differentiable under your definition. – user127096 Mar 14 '14 at 22:47

1. On the "continuous" side, a map $f: X\to R^n$ defined on an open nonempty subset $X\subset R^m$ is continuous if and only if its graph $\Gamma_f$ is an $m$-dimensional topological manifold. (This is an immediate consequence of Brouwer's open mapping theorem applied to the projection $\Gamma_f\to X$. This, of course, will not help you with a "simpler proof" of the inverse mapping theorem!) For compact $X$, you can equivalently define continuity by requiring graph to be compact.