Continuous non-vanishing vector field on annulus shape regions

Question

Let $\Omega \subset \mathbb{R}^2$ be an open bounded region and $\Omega_0 \subset \subset \Omega$. Assume also that $\Omega_0$ is simply-connected. Suppose a continuous non-vanishing vector field $F$ is defined on $(\Omega \backslash \Omega_0 )^c$. Can $F$ be extended to a continuous non-vanishing vector field $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$?

Answer 1

I am unsure of exactly what your hypotheses are supposed to be, but does this give a counterexample?

Take $\Omega_0 = \{x\in\mathbb R^2: 1<\|x\|<2\}$ and $\Omega= B(0,2) = \{x \in \mathbb R^2: \|x\|<2\}$ and $F(x_1,x_2) = \frac{1}{\sqrt{x_1^2+x_2^2}}(x_2,-x_1)$. The $\Omega\backslash \Omega_0 = \{x\in \mathbb R^2: \|x\|\leq 1\}$, and $F$ is defined on all of $\{x\in \mathbb R^2: \|x\|>1\} = (\Omega\backslash \Omega_0)^c$, but clearly $F$ cannot be extended continuously to a function $ \tilde{F}\colon \Omega \to \mathbb R^2$ since $\lim_{x \to 0} F(x)$ does not exist.

Update

For any $a \in \mathbb R^2$ let $F_a(x)= F(x-a)$ where $F(x_1,x_2) = \frac{1}{\sqrt{x_1^2+x_2^2}}(x_2,-x_1)$ is as above. Then the vector field $F_a$ is smooth on $\mathbb R^2 \backslash \{a\}$ but cannot be extended to a continuous vector field at $a$. It follows that if $\Omega_0 \subset\subset \Omega$ and we pick $a=(a_1,a_2) \in \Omega\backslash\Omega_0$, then the vector field $F_a(x)$ is defined on $(\Omega\backslash \Omega_0)^c$ and cannot be extended to all of $\Omega$.