Let $f:\mathbb{S}^1\to\mathbb{R}^2\setminus\{0\}$ be a smooth function. Prove that there exists a continuous $\hat{f}:\bar{\mathbb{D}}^2\to\mathbb{R}^2\setminus\{0\} \Leftrightarrow \text{deg}(f)=0$. Furthermore, prove that the extension can be chosen smoothly.
I just have a vague hunch: if there is such a $\hat{f}:\bar{\mathbb{D}}^2\to\mathbb{R}^2\setminus\{0\}$, then $\hat{f}$ is homotopic to a constant, since $\bar{\mathbb{D}}^2$ is contractible, so $\text{deg}(\hat{f})=0$, and I'd like to conclude that $\text{deg}(f)=0$.
For the opposite direction, if $\text{deg}(f)=0$, that means the image of $f$ is not a closed curve, but a compact curve with both ends unconnected, which is diffeomorphic to a closed interval. So we have an induced smooth function $f:\mathbb{S}^1\to[a, b]$, so we can define $\hat{f}(x):=||x||f\left(\frac{x}{||x||}\right)$ for $x\neq 0$ $\hat{f}(0):=0$, which is continuous.
I don't know if this works, but that was the only thing I could think about. Any tips? Thanks!
The idea for the first direction basically looks fine. The second part does not really make sense, because the image of $f$ always is a closed curve (also you have to construct $\hat f$ with values in $\mathbb R^2\setminus\{0\}$.
As a hint for the solution, observe that for any space $Y$ a continuous extension of $f:S^1\to Y$ to $\hat f:\bar D^2\to Y$ is equivalent to a homotopy $H:S^1\times [0,1]\to X$ between $f$ and the constant map to $\hat f(0)\in Y$. The relation simply is that $H(x,t)=\hat f((1-t)x)$.