Let $f:\Bbb{R}^2\to\Bbb{R}^2$ be a continuous map such that there exists $r>0$ with $f(x)\in B(x,r)$ for every $x\in\Bbb{R}^2$. Show that $f$ is surjective.
I have some vague idea that if $p\not\in f(\Bbb{R}^2)$, one could somehow construct a retraction from $\Bbb{R^2}$ to $S^1$ via a homeomorphism between $\Bbb{R}^2\setminus\{p\}$ and $S^1$ so that one has a contradiction. I don't really see how to do it. Also is there a way to do it without explicitly using methods in algebraic topology?
If you allow homotopies to be used, then there is a way to do it without algebraic topology.
Just take the homotopy $$ H\colon \mathbb{R}^n\times I \to \mathbb{R}^n, \ (x,r) \mapsto rf(x)+ (1-r)x .$$ from $f$ to the identity of $\mathbb{R}^n$. Then $H(x,i)\in B(x,r)$ for all $x\in \mathbb{R}^n$ and $i\in I$. This condition ensures that we can extend this homotopy to $\mathbb{S}^n$.
It is easy to show that a non surjective self map of the sphere is null-homotopic. So if $f$ is not surjective, then the identity of $\mathbb{S}^n$ is null-homotopic.
If you use an analytical proof of Brouwer's fixed point theorem to show that the sphere is not contractible, you have proven that $f$ is surjective without using homology or homotopy groups.
