How to prove that a space is not a differential manifold?

Question

Given a box （the surface of a cubic） in R^3 space, can I give a smooth structure on it to make it a differential manifold?

Answer 1

Yes. The box $B$ is homeomorphic to the $2$-sphere $S^2$, so you can choose a smooth structure on $S^2$ and a homeomorphism $f: B \to S^2$ and use that to transfer all your charts: a chart $\varphi : S^2 \supseteq V \to U \subseteq \mathbb{R}$ becomes $\varphi \circ f$.

Transferring a smooth structure will result in a smooth structure, since if a transition function $\varphi \circ \psi^{-1}$ is smooth, then clearly $(\varphi \circ f) \circ (\psi \circ f)^{-1} = \varphi \circ \psi^{-1}$ is also smooth.