# 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?

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PLEASE AVOID CAPITAL LETTERS. THIS IS CONSIDERED AS SHOUTING. IT IS UNNECESSARY TO SHOUT, WE CAN ALL HEAR YOU JUST FINE. – Asaf Karagila Oct 14 '12 at 7:37
Please put in at least a minimal effort at formatting your post (specifically, punctuate and use the Shift key appropriately). This sort of thing is rather disrespectful. – Cameron Buie Oct 14 '12 at 7:40
@Asaf: ssshhhhh..... – arjafi Oct 14 '12 at 7:44
I do not understand what $f$ and $g$ have to do with the final question. – PAD Oct 14 '12 at 7:52
Why did this question get down voted? It's a good question. – Rudy the Reindeer Oct 15 '12 at 18:05

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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Sorry, would you elaborate on the last sentence, please? How do you use Brouwer's fixed point theorem? Thank you. – Rudy the Reindeer Oct 14 '12 at 12:16
@MattN.: OK, I hope it's clearer now, if not, comments are welcome. – Dejan Govc Oct 14 '12 at 12:53
@MattN.: I'm mentioning it simply because the question is mentioning it. I think the questioner is interested in the following question: "Do there exist continuous functions $f,g:D^2\to D^2$, such that $f$ and $g$ are surjective and for all $z\in D^2$ we have $f(z)\neq g(z)$?" The questioner then proposes using Brouwer's fixed point theorem to derive a contradiction from the existence of such two functions, which would show that such functions cannot exist. I am showing the contrary: that such functions do indeed exist, and thus Brouwer's theorem cannot be used to disprove their existence. – Dejan Govc Oct 14 '12 at 13:16
@MattN.: You're right, the question doesn't explicitly mention it. But I think the author was implicitly suggesting that the existence of such functions might contradict Brouwer's theorem. I might be mistaken, but I cannot imagine any other reason for mentioning that retraction. By the way, I think the easiest way to visualize the functions is by imagining $y$ as time. This way you can see $F$ and $G$ as two paths moving in the square, trying to avoid each other at all times. – Dejan Govc Oct 15 '12 at 18:38
@MattN.: Anytime. =) And yes, your understanding of this thread is the same as mine. – Dejan Govc Oct 15 '12 at 18:47

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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uhhh...i don't think this answers the question :p, i.e. i think the result he wants is that any two surjections from the disk to itself have to agree somewhere, and what you're writing is the how to finish the contradiction he's going for – uncookedfalcon Oct 14 '12 at 7:49
yes thats right – jim Oct 14 '12 at 7:54
@uncookedfalcon I'm sorry, could you elaborate what you mean? The question seems to ask "Can I construct a retract of the disk?" and the answer says "No, if you could we'd have the following contradiction." – Rudy the Reindeer Oct 14 '12 at 8:21
@MattN.: The question is asking if it is possible to construct such a retraction given functions $f$ and $g$ with the properties stated. Now if two such functions don't exist, then being in possession of them is a contradiction and from a contradiction anything follows, for example, that the retraction exists (and so does Santa Claus). – Dejan Govc Oct 14 '12 at 8:37
Dear @DejanGovc, thank you very much! Now I understand! : ) – Rudy the Reindeer Oct 14 '12 at 8:40