Suppose \(f: \mathbb{R} \to \mathbb{R}^2\) and \(g: \mathbb{R}^2 \to \mathbb{R}\) are differentiable. Prove that the composite function \(f \circ g\) cannot have a differentiable inverse.
Solution
Let \[ h=f\circ g:\mathbb R^2\to\mathbb R^2. \] For every \(x\in\mathbb R^2\), the chain rule gives \[ Dh_x=Df_{g(x)}\,Dg_x. \] Here \(Dg_x:\mathbb R^2\to\mathbb R\) has rank at most \(1\), so \(Dh_x\) has rank at most \(1\). In particular \(Dh_x\) is never an invertible linear map on \(\mathbb R^2\).
If \(h\) had a differentiable inverse \(k\), then \[ k\circ h=\operatorname{id}_{\mathbb R^2}. \] Differentiating at \(x\) would give \[ Dk_{h(x)}\,Dh_x=I_2, \] which would force \(Dh_x\) to be invertible, a contradiction. Therefore \(f\circ g\) cannot have a differentiable inverse.