This is a question in the context of classical differential geometry. We define a patch to be a differentiable function $f$ from an open subset $A$ of $\mathbb R^2$ to $\mathbb R^3$ which is also a homeomorphism from $A$ onto its image (which we call the trace of $f$). A differentiable surface is defined to be a subset $S$ of $\mathbb{R}^3$ such that for each $p\in S$ there's an open neighborhood $V$ of $p$ in $\mathbb{R}^3$ such that $S\cap V$ is the trace of a patch.
It seems obvious that the boundary of a cube is not a differentiable surface in this sense because of the edges and corners, but how does one prove this rigorously?