Why do we need null-homotopy to follow that a vector field is conservative if its curls is zero?

Question

so let $\vec{v}: \Omega \subset \mathbb R^n \to \mathbb R^n$ be a vector-field.

Now we know:

(I) $\vec{v}$ conservative $\quad\Rightarrow\quad \nabla\times\vec{v}=0$

If further, $\Omega$ is simply-connected i.e. path-connected and null-homotopic, we can also state:

(II) $\nabla\times\vec{v}=0 \quad \Rightarrow \quad \vec{v}$ is conservative.

Question: Why do we need null-homotopy for (II)?