How do I prove that the boundary of the unit square is not a $\mathscr{C}^1$ manifold?

Question

We are to prove that $\partial([0,1]^2) \subset \mathbb{R}^2$ ie the boundary of the unit square is not a $\mathscr{C}^1$ manifold.

Currently, here's how I'm thinking about this but am unable to put it down 'rigorously'.

$X \subset \mathbb{R}^N$ (for some $N \in \mathbb{N}$) is a $\mathscr{C}^k$ manifold if there exists a $\mathscr{C}^k$ diffeomorphism between $X$ and and open set $U \subset \mathbb{R}^k$. That is, we have a $\mathscr{C}^k$ parametrization (classical smooth map)between $U$ and $V \text{ open}\subset X$ about any point in $X$ and a $\mathscr{C}^k$ coordinate system (not a classical smooth map) in the other direction.

In the case of our square, looking at our edges, we can find such diffeomorphisms. However, when we consider our vertices, we cannot find a differentiable function on any open set containing a vertex since the derivative at the vertex does not exist.

Is this on the right track?

Answer 1

HINT: Think about the parametrization rather than the chart. Say $0\in U\subset\Bbb R$ maps to the corner under $\phi$. What is the image of $d\phi_0$? Now think about continuity of the derivative as you approach $0$ from the left and the right.