Proposition

If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, the graph of $f$ is $\{ (x, y) \mid y = f(x) \}$. Show that the graph of $f$ is an $n$-dimensional manifold if and only if $f$ is differentiable.

Solution

Suppose $f$ is differentiable.

Let $g: \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}^m$ be defined such that $g(x, y) = y - f(x)$. Since $f$ is differentiable, $g$ is differentiable. Moreover, $Dg(x, y) = \begin{bmatrix} -D_xf(x) & I_m \end{bmatrix}$, so it has rank $m$. By Theorem 5-1[Spivak], $g^{-1}(0) = \{ (x, y) \mid y = f(x) \}$ is an $(m + n - m)$-dimensional manifold.

The converse is not true as discussed here.