Showing $0$ is in the image of this function I want to show that any continuous function $f:\;\mathbb{R}^2\to\mathbb{R}^2$ with $||f(x)-x||\leq M$ for some $M>0$ has $0$ in its image. 
The hint is to find a function $g:\;\mathbb{R}^2\to\mathbb{R}^2$ which sends a unit disc into itself, such that $g(x)=x$ implies $f(x)=0$. We can then use Brouwer's fixed point theorem to deduce that there is such an $x$.
I can't see how to construct $g$ though? The obvious choice $f(x)=g(x)+x$ doesn't seem to give me the disc-into-disc property. Any ideas/further hints appreciate.
 A: By contradiction:
Pick $d$ "large enough" and let $i$ be the inclusion from the circle $C_d$ with radius $d$ to the closed ball $B_d$ of radius $d$ (centered at the origin).
Consider the continuous map $g:B_d\rightarrow C_d$ defined by $g(x)=df(x)/|f(x)|$.
Consider the map $C_d\overbrace{\rightarrow}^i B_d\overbrace{\rightarrow}^{g} C_d$.
Now consider the induced homomorphisms on the fundamental groups. Since $B_d$ is simply connected the composition of the induced homomorfisms must be the trivial homomorphism. In other words if we take the loop consisting one trip around the circle this must map to a trivial loop. Clearly if $d$ is larger than $M$ this will be impossible.
Just to make this clear we can show that a loop $l$ is homotopic to the loop $(i\circ g)_*l$. We need a homotopy $H:C_d\times I\rightarrow C_d$ such that $H$ is $l$ in $C_d\times \{0\}$ and $(g\circ i)_*l$ in $C_d\times \{1\}$. All you have to do to define $C_d(x,t)$ is consider the small ark between $x$ and $g(x)$ (this is well defined because $d>M$ and so the point cannot move across the circle), to prove the function is continuous it suffices tu use the $\epsilon-\delta$ definition.
