Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we have a set $X$ with $X=U \cup V$. If we pick a permutation $f$ of $U$ and a permutation $g$ of $V$ which agree on the intersection $U \cap V$, we can coalesce them into one big endo-map $F$ of $X$. In general, of course, $F$ is no longer a permutation - we can only say that it is onto. A counter-example is fun to think of - for example, if $X$ is finite, then $F$ is always a permutation.

Now let $X, U$ and $V$ be open subsets of $\Bbb{R}^n$, and suppose $f$ and $g$ are smooth homeomorphisms. I have been unable to come up with a single example for which the above patching fails "badly", say for which the set of points $x$ having more than one pre-image had positive measure.

I'd love to know if anyone knows anything related to this problem, or could provide a counter-example. Thank you!

share|improve this question
    
This may turn out to be useless for your situation, but then again it may not. mathoverflow.net/questions/30661 I suggest it because it might inspire a construction that is as bad as you want. I am not clear on how bad "bad" is for you. Gerhard "Ask Me About System Design" Paseman, 2011.11.02 –  Gerhard Paseman Nov 3 '11 at 4:40

2 Answers 2

up vote 4 down vote accepted

If you allow the open sets $X$, $U$ and $V$ to be disconnected, you get a counter-example by taking a countable example and changing points to pairwise disjoint balls.

Edit 2: Here is a modified version of Joel's example, which was later deleted. Define $U$ to be the open set inside the green rectangle and $V$ to be the open set inside the red rectangle. Just for notation, suppose that the left-bottom corner is $(0,0)$ and that the width of the "house" is $1$.

Define $f \colon U \to U$ be a mapping $$(x,y) \mapsto \left(\frac12x,y\right)$$ on the part of $U$ which does not include the "chimney", i.e. the top left box. Correspondingly, the part of the mapping $g \colon V \to V$ which maps everything but the chimney is defined as $$(x,y) \mapsto \left(\frac12(x+1),y\right).$$ The chimneys are then mapped with homoemorphisms to the missing parts of $U$ and $V$. Now the set $U \cap V$ is the middle part of the house. The set inside the yellow rectangle has two pre-images and has positive area.

Remark: This example does not differ that much from the first one which I posted. Essentially we just extend the mappings defined on the balls slightly (and rearrange the balls if this is not possible otherwise) to make the sets $U$ and $V$ connected.

share|improve this answer
    
Thank you very much Tapio for your answer. It seems Joel has deleted his answer, so I'm having trouble piecing your description together. Would you mind adding just a couple of extra details before I accept your answer as definitive? Thank you! –  Bruno Joyal Nov 8 '11 at 22:54
    
Bruno, you are welcome! I now edited the answer. I prefer not to write an explicit formula for the whole mappings $f$ and $g$. Hopefully this amount of detail satisfies you. –  Tapio Rajala Nov 9 '11 at 7:36
    
It is great, thank you again :-) –  Bruno Joyal Nov 10 '11 at 3:53

The set of points having more than one pre-image is open (so it always has positive measure if it is non-empty). It follows that $U \cap V$ must be disconnected and in fact it has to have infinitely many connected components. However there are examples in $\mathbb{R}^2$ where both $U$ and $V$ are connected:

For a negative integer $n$ we define the real intervals $$I_n=(2n-\frac{1}{3},2n+\frac{1}{3}) \mbox{ ; } J_n=(2n+\frac{2}{3},2n+\frac{4}{3}).$$ For a non-negative integer $n$ we let $$I_n=J_n=(n-\frac{1}{3},n+\frac{1}{3}).$$

Let $f$ and $g$ be two increasing (smooth if you wish) homeomorphisms from $\mathbb{R}$ onto $\mathbb{R}$ such that $f[I_n]=I_{n+1}$ and $g[J_n]=J_{n+1}$ for all $n \in \mathbb{Z}$. Let $I=\bigcup_{n\in \mathbb{Z}}I_n$ and $J=\bigcup_{n\in \mathbb{Z}}J_n$.

Now in $\mathbb{R}^2$ we define the open sets $$U=I \times (-1, \infty) \cup \mathbb{R} \times (2,\infty)$$ $$V=J \times (-\infty, 1) \cup \mathbb{R} \times (-\infty ,-2).$$

Finally we let $F=f \times id$ and $G=g \times id$ be the autohomeomorphisms of $U$ and $V$ respectively. The set of points with two pre-images is precisely $I_0 \times (-1,1)$.

Similar examples exist in $\mathbb{R}^n$ for $n\geq3$, but of course in $\mathbb{R}$ we cannot get $U,V$ to be connected.

share|improve this answer
    
Thank you for this beautiful example, Ramiro! –  Bruno Joyal Nov 8 '11 at 22:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.