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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?

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    Why did this question get down voted? It's a good question.2012-10-15

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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.

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    @DejanGovc How about constructing arrows in disc $D$ with tail at $f(d)$ and head at $g(d)$ for all $d \in D$, and every arrow points to some point on the boundary of disc. Now, we can construct$a$map from $f(d)$ to some point on the boundary of $D$ by following the arrows. Since $f$ is surjective, every $d \in D$ is mapped to some point on boundary. Suppose this new map is $a:D\rightarrow C$ where $C$ is boundary. Now, I ask if $a$ is continuous and surjective. If yes, then $a$ is the needed retraction. Please see axiom 2 in https://math.stackexchange.com/q/2496984/3437012017-11-14
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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 $H_1(S^1) = \tilde{H}_1(S^1) = \Bbb{Z}$.

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    @BenjaLim The question is asking whether, given two continuous functions $f,g: D^2 \twoheadrightarrow D^2$ such that $f(z) \neq g(z)$ for all $z \in D^2$, it is possible to construct a retraction of the disk to its boundary $S^1$.2012-10-15