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?
Retraction map from unit disk to its boundary
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7**PLEASE AVOID CAPITAL LETTERS.** THIS IS CONSIDERED AS SHOUTING. IT IS UNNECESSARY TO SHOUT, WE CAN ALL HEAR YOU JUST FINE. – 2012-10-14
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2Please 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. – 2012-10-14
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2I do not understand what $f$ and $g$ have to do with the final question. – 2012-10-14
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2Why did this question get down voted? It's a good question. – 2012-10-15
2 Answers
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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0Sorry, would you elaborate on the last sentence, please? How do you use Brouwer's fixed point theorem? Thank you. – 2012-10-14
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1@MattN.: OK, I hope it's clearer now, if not, comments are welcome. – 2012-10-14
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0Thank you very much for the edit. Your train of thought is very clear. I don't understand though why you mention Brouwer's fixed point theorem. I know that in its proof there is a construction of a retraction. And we know that there is no retraction $D^2 \to S^1$ hence we cannot have $f(z) \neq z$ for all $z$. What I still haven't managed is to understand where this comes in when we have two functions. I'm sorry for being so slow but would you please elaborate this one detail for me? Many thanks! – 2012-10-14
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2@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. – 2012-10-14
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0Thank you very much for your reply! I will need to think about this for some time. – 2012-10-14
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0I'm sorry I still don't understand the sentence about the fixed-point theorem. Isn't the argument just "No we cannot construct a retraction since such functions exist but we know that a retraction $D^2 \to S^1$ doesn't"? (You wrote that the question mentions Brouwer but I don't see where it does.) I'm sorry for being so slow. – 2012-10-15
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0I'm impressed how you came up with these functions! Now I shall spend some time to see what they do. – 2012-10-15
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1@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. – 2012-10-15
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0Thank you for helping me "see" the functions! : ) (I thought the retraction was mentioned because we know such a retraction does not exist hence if we can use $f$ and $g$ to construct such a retraction we have shown that such functions cannot exist. At least that's my current understanding of this thread.) – 2012-10-15
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1@MattN.: Anytime. =) And yes, your understanding of this thread is the same as mine. – 2012-10-15
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0Nice! : ) Thank you for confirming that! : ) – 2012-10-15
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0@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/2496984/343701 – 2017-11-14
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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0uhhh...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 – 2012-10-14
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0yes thats right – 2012-10-14
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0@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." – 2012-10-14
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2@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). – 2012-10-14
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0Dear @DejanGovc, thank you very much! Now I understand! : ) – 2012-10-14
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0$Af$ and $g$ are homotopic (where $A $ is the antipodal map) in an obvious way. Then you apply this result to $f=f$ and $g=id$. – 2012-10-14
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0@uncookedfalcon It is unclear to me what the question even is. I thought he was asking if a retract of the disk onto $S^1$ exists. – 2012-10-14
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0@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