Retraction map from unit disk to its boundary Given two continuous surjective functions $f$ and $g$ from the unit disk to itself and $f(z) \neq g(z)$ for all $z$ in the unit disk is it possible to construct a retraction map from the unit disk to its boundary?
 A: There is no retract from $D^2$ to $S^1$. If there were such a retract $r : D^2 \to S^1$ then we would have an injection
$$i_\ast : H_1(S^1) \to H_1(D^2)$$
where $i_\ast$ is the map induced from inclusion $i : S^1 \to D^2$. This is because $r \circ i =1$ and hence on homology $r_\ast \circ i_\ast = 1$, viz. $i_\ast$ has a left inverse. But this is impossible because $H_1(D^2) = \tilde{H}_1(D^2)= 0$ while $H_1(S^1) = \tilde{H}_1(S^1) = \Bbb{Z}$.
A: No, it's not possible. Let $X=[0,1]\times[0,1]$ and define $F:X\to X$ by the formula $$F(x,y) =\begin{cases}
\left(\frac53 x, y\right);& x\in\left[0,\frac15\right]\\
\left(\frac13,(2-5x)y\right);& x\in\left[\frac15,\frac25\right]\\
\left(\frac13,(5x-2)y\right);& x\in\left[\frac25,\frac35\right]\\
\left(\frac53(2x-1),y\right);& x\in\left[\frac35,\frac45\right]\\
\left(1,y\right); &x\in\left[\frac45,1\right]
\end{cases}$$
and $G:X\to X$ by
$$G(x,y) =\begin{cases}
\left(1-\frac53 x, y\right); &x\in\left[0,\frac15\right]\\
\left(\frac23,1+(2-5x)(y-1)\right); &x\in\left[\frac15,\frac25\right]\\
\left(2-\frac{10}3 x,1\right); &x\in\left[\frac25,\frac35\right]\\
\left(0,1-(5x-3)y\right); &x\in\left[\frac35,\frac45\right]\\
\left(\frac{10}3 x-\frac83,1-y\right); &x\in\left[\frac45,1\right]
\end{cases}$$
It is an easy exercise to verify that $F$ and $G$ are both well-defined (and thus continuous), surjective and $F(x,y)\neq G(x,y)$ for all $(x,y)\in X$.
Now, just choose a homeomorphism $\phi:D^2\to X$, for example $$(x,y)\mapsto \frac12\frac{\|(x,y)\|_2}{\|(x,y)\|_\infty}(x,y)+\left(\frac12,\frac12\right)$$ will do. Define $f=\phi^{-1}\circ F\circ\phi$ and $g=\phi^{-1}\circ G\circ\phi$. These functions $f$ and $g$ inherit their properties from $F$ and $G$ and are thus continuous, surjective and $f(z)\neq g(z)$ for all $z\in D^2$.
This shows that functions $f$ and $g$ with properties from the question do indeed exist. This means that the existence of such functions is not contradictory, and as such also cannot contradict Brouwer's fixed point theorem. Hence, we cannot deduce a contradiction (=existence of a retraction) from the existence of such functions.
