## Help in Formalizing a proof using liftings

Hi,

The question is: If $g:\mathbb{R}^2\to \mathbb{R}^2$ is a continuous map and there is $r>0$ such that $g(x)\in B(x, r)$ (ball center $x$, radius r) for every $x\in \mathbb{R}^2$, then $g(\mathbb{R}^2)=\mathbb{R}^2$.

I have an idea of a proof but I'm not sure how to formalize it using liftings, etc (I'm not well-practiced in doing calculations with non-trivial fundamental groups). After supposing there is $a$ in $\mathbb{R}^2$ such that $g(\mathbb{R}^2)\subseteq \mathbb{R}^2-\{a\}$, my idea is to consider the standard loop in $\mathbb{R}^2$, $e^{2\pi it}, \, t\in[0,1]=:I$, but with radius $2r$ and centered at $a$, call it $\alpha$. Each point $\alpha(t)$ has image under $g$ contained in $B(\alpha(t), r)$, and so $g(\alpha(I))$ is contained in the annulus, center a, with inner radius r, outer radius 3r. Put $\beta=g\circ \alpha$. Since $g$ is continuous and $\alpha$ is homotopic to the constant loop at $a$, $\beta$ should be homotopic to a constant loop. But joining points of $\beta$ in the balls around the four axes-intercepts $\alpha(0),\, \alpha(1/4),\, \alpha(1/2), \, \alpha(3/4), \, \alpha(1)$, for example, (and perhaps winding around again to be certain), we can see that $\beta$ has to wind around $a$. So since $\beta(I) \subseteq \mathbb{R}^2-\{a\}$, it isn't homotopic to a constant loop, a contradiction.

It's frustrating that I have the idea but I am not not familiar enough with performing calculations to formalize this. If someone could help me get started I'd be most appreciative!

Many thanks