# Defining a function via restrictions of many bijections

I'm trying to define a certain surjective map from $(0,1) \to (0,1)$. For any $(a,b) \subseteq [0,1]$, there's a bijection $(a,b) \rightarrow (a,b) \cup (\frac{1}{3},\frac{2}{3})$. Is there a way to define these bijections such that they're all "compatible"? Basically I'm searching for a well-defined function that on intervals of the above form has the same behavior as the above? Thank you for any help/advice!

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The bijections would have to be of various types, since the union $(a,b) \cup (\frac13,\frac23)$ might be two intervals, or a single interval, and the relative positions of $a,b,\frac13,\frac23$ determine this. It also is unclear what you are asking, is it for a single map $f:[0,1]\to (0,1)$ whose restrictions to each of the intervals formed by $a,b$ is a bijection of the form above? –  coffeemath Aug 2 '13 at 20:16
@coffeemath Thanks for your comment. Yes, I'm looking for a single function. I've also simplified the question a bit - i've decided that I'm not so concerned with the endpoints. –  Vien Nguyen Aug 2 '13 at 20:30
I think "compatible" is really vague here. –  Tunococ Aug 2 '13 at 21:38
@Tunococ Thanks for asking for clarification. What I mean is that the function $f: (0,1) \rightarrow (0,1)$ that I'm looking for, when restricted to any $(a,b)$, should be exactly the bijection $(a,b) \rightarrow (a,b) \cup (\frac{1}{3},\frac{2}{3})$. My questions is basically "Are there ways of choosing these bijections $(a,b) \rightarrow (a,b) \cup (\frac{1}{3},\frac{2}{3})$ such that the above is true? –  Vien Nguyen Aug 2 '13 at 22:13
Definitely there is not. You can first pick $(a, b) = (0, 1)$. That defines the whole function. Then it is impossible that the restrictions of that function to $(0, 1/3)$ and $(2/3, 1)$ are bijections. –  Tunococ Aug 2 '13 at 22:25

Let $I=(\frac13,\frac23)$ and $J=(a,b)$, and assume $f$ is the map from $(0,1)$ to itself which gives a "compatible choice" for all the bijections, by which is meant that the restriction of the single map $f$ to the domain of each suppoped bijection gives that bijection.

The requirements then are that:

[1] For any $J$ the map $f:J \to J \cup I$ be a bijection.

Now note that $I \cup I=I.$ so taking $J=I$ in [1] gives that $f:I \to I$ is a bijection.

Next take any interval $K$ which is a proper subset of $I$, and note that $K \cup I=I$, so that applying [1] with $J=K$ gives that $f:K \to I$ is a bijection. This is a contradiction, since $K$ is a proper subset of $I$ and $f$ is already a bijection from $I$ to $I$.

To bring out the contradiction: Choose $x$ in $I$ but not in $K$. Then $f(x) \in I$, so that since $f:K \to I$ is onto there is $y \in K$ for which $f(y)=f(x)$. This contradicts that $f$ is a bijection on $I$, since $x \neq y$ by our choice.

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Thank you @coffemath ! I'll keep this in mind for later. –  Vien Nguyen Aug 2 '13 at 22:18
@VienNguyen It seems interesting why you were trying to find such $f$ with those bijections. Maybe if you post another question covering the job these bijections were trying to do, someone could give help on reformulating some similar question which had a positive answer. –  coffeemath Aug 2 '13 at 22:35
This came out of a question that my undergrad research advisor asked me to think about, i.e. whether or not there are open maps from [0,1] to the real line. Keeping in mind that (0,1) and the real line are homeomorphic, my idea was to map 0 and 1 separately into (1/3,2/3) here, but that didn't seem to work out XD. I think I want to think about this some more before asking the MSE community. But thank you for the help! –  Vien Nguyen Aug 5 '13 at 2:55

If (1/3, 2/3) is a proper subset of (a, b), then use the identity map.

Otherwise, divide the domain interval (a, b) into two parts, mapping the first part onto (a, b) using scaling, and mapping the second part onto (1/3, 2/3) - (a, b) using scaling and translation.

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This just shows for any $(a,b)$ there is such a bijection, but does not settle whether all such may be chosen "compatibly" as asked in the OP. –  coffeemath Aug 2 '13 at 22:00
Fair comment. I guess what is needed is to define "compatible functions". Plainly not all bijections would be "compatible" in the naive sense of the word. E.g., a continuous bijection would presumably not be compatible with a non-continuous bijection. –  Nick R Aug 2 '13 at 22:15
@NickR thanks for helping out, and sorry I wasn't more clear with defining "compatible". coffeemath's definition in his/her answer is what I meant. –  Vien Nguyen Aug 2 '13 at 22:20