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We are given a (closed) ball $D^n$ and a (continuous) map $f: D^n \to D^n$, that is identity on the boundary of $D^n$.

Let $C$ be a subset of $D^n$, and let $f^{-1}(C)$ be the inverse image of $C$ in $D^n$.

The claim is that there exists a map $g: C \to f^{-1}(C)$ that is identity on the intersection of $C$ and the boundary of $D^n$.

I actually suspect that the claim is wrong in general, but cannot find a counter-example. Also, if it is indeed wrong, what are the conditions on $C$ and $f$ so that it is correct?

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  • $\begingroup$ I edited your post so that the maths works. Have a look at the source (by clicking 'edit') to see what I did. $\endgroup$
    – David Roberts ♦
    Commented Dec 1, 2011 at 8:56
  • $\begingroup$ You probably require $g$ to be continuous... $\endgroup$
    – Malik Younsi
    Commented Dec 1, 2011 at 14:29
  • $\begingroup$ Sure, all maps here are continuous. $\endgroup$
    – user19747
    Commented Dec 5, 2011 at 8:42
  • $\begingroup$ I've spent some time thinking about it and there are probably a lot of different cases to study (If I am not missing something). Can you specify what you are looking for, or do you really need the answer in its most general form? $\endgroup$
    – Simon Markett
    Commented Dec 5, 2011 at 11:08
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    $\begingroup$ Take $B$ to be the union of $a\sin\frac{\pi}x$ ('$0<x\le 1$'), interval from $-1$ to $0$ and the interval from $-ia$ to $ia$ on the complex plane. Now take any continuous map $f$ that is identity on the boundary, sends $[-ia,ia]$ to one point and doesn't glue any other two points together. Put $C=f(B)$. Then $C$ is a continuous path from $-1$ to $1$ but $B=f^{-1}(C)$ contains no such path. $\endgroup$
    – fedja
    Commented Dec 6, 2011 at 22:25

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This question is a real mess, due to all the posts by Petr before he figured out about comments. But if you read it through carefully, fedja has provided an answer, which I am copying here so this question will stop being bumped to the front page. The point is that the answer is no. We'll produce a set $C$ which will be a counterexample.

We start by defining a subset of $\mathbb{C}$. Let $A = [-1,0) \cup (0,1] \subset \mathbb{C}$. Fix a real number $0 < a < 1$, and consider the curve $x\mapsto x+ia\sin(\pi/x)$. Because $0 < a < 1$, we know this curve stays inside the unit disk. Define

$$B = \cup_{x \in A} (x+ia\sin(\pi/x)) \cup [−ia,ia]$$

Now take any continuous map $f$ that is identity on the boundary, sends $[−ia,ia]$ to one point (zero, for instance), and otherwise doesn't send any two points to one point (i.e. is injective away from $[-ia,ia]$). It is easy to construct such an $f$. Put $C=f(B)$. Then $C$ is a continuous path from $−1$ to $1$ but $B=f^{-1}(C)$ contains no such path, i.e. there is no continuous map $g$ from $[-1,1]$ to $B$ which is the identity at $-1$ and $1$. Therefore the claim is false.

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  • $\begingroup$ Now I hope Petr comes back and deletes his answers which should have been comments. Then this will whole thread will look much better to future viewers. But at least this way, the answer will be clear. $\endgroup$
    – David White
    Commented Feb 16, 2012 at 21:45

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