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