There is no smooth diffeomorphism $f:\mathbb R^3 \setminus [-\frac{1}{2}, \frac{1}{2}]^3 \to \mathbb R^3$

Question

I want to prove thtat there is no smooth diffeomorphism $f:\mathbb R^3 \setminus [-\frac{1}{2}, \frac{1}{2}]^3 \to \mathbb R^3$.
Attempt:
I have seen a lot about $\mathbb R^3 \setminus \{0\}$, I don't know if here it works the same way, since the hole is of similar type, but not just one point. We know that the second de Rham cohomology group of $\mathbb R^3$ vanishes: $$H^2(\mathbb R^3) = \{0\}.$$ However, take the $2$-form $\omega $ defined by $$\omega_{(x,y,z)} := \frac{x\, dy \wedge dz + y \,dz\wedge dx + z \,dx\wedge dy}{(x^2+y^2+z^3)^{3/2}}$$ It is a quick calculation that $\omega$ is closed, i.e. $\omega \in Z^2(\mathbb R^3\setminus[-\frac{1}{2}, \frac{1}{2}]^3). $ But since $$\int_{S^2} \omega = 4\pi \neq 0,$$ $\omega$ is not exact ($[-\frac{1}{2}, \frac{1}{2}]^3 \subseteq B_1(0)$). Now if there was a smooth diffeomorphism $f$, $\omega$ pullbacked with $f^{-1}$ would be a non-exact form too since the cohomology classes are the same. Is this correct?

Answer 1

I'll elaborate my comment into an answer since it's getting longer.

Yes, you're right. The space you're considering, let's call it $A$, is homeomorphic to $\Bbb R^3 \setminus \{ 0\}$, hence no homeomorphism exists between $A$ and $\Bbb R^3$. This also implies that there is no diffeomorphism between these two spaces.

Exhibiting a 2-form like you did is also right, though what I was saying is that being the two spaces homotopically equivalent, their De Rham groups are going to be isomorphic.