3
$\begingroup$

In another MathOverflow post I asked: In a topological space if there exists a loop that cannot be contracted to a point does there exist a simple loop that cannot be contracted also?

Note that once consequence of this is that for spaces in which this holds you only need to consider simple loops when proving that a space is not simply connected. This also suggests that looking at path-connected spaces only would still be interesting.

Some very elegant and interesting counterexamples were given, both Hausdorff (for a space embedded in $\mathbb{R}^3$) and non-Hausdorff.

I propose a follow up question which is to determine the spaces with the property that if there exists a loop that cannot be contracted to a point there exists a simple loop that cannot be contracted also?. (Or equivalent to classify the counterexamples in some way)

For example what conditions can be place on a space embedded in $\mathbb{R}^n$ to force the condition to hold?

Update: Given that this is quite a broad question I have asked another question specifically about the $\mathbb{R}^2$ case: In a subset of $\mathbb{R}^2$ which is not simply connected does there exist a simple-loop that contracts to a point?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
-2
$\begingroup$

Some very sketchy and preliminary thoughts of my own that I didn't want to place in the question to reduce clutter:

I believe that this property does hold unconditionally for any subset of the plane, $\mathbb{R}^2$. I gave an elementary argument for this in the previous post which may not be correct but I think is plausible. (A slick proof of this or counterexample would be great).

Based on the $\mathbb{R}^3$ counterexample which consisted of two connected spaces joined at a single point I venture the highly speculative guess guess that for any space embedded in $\mathbb{R}^n$ the condition does hold if the space cannot be separated by the removal of a finite number of points. I have no proof for this but I think that my speculative proof for $\mathbb{R}^2$ may work for these cases as without such points you can't force simple loops to stay in simply connected components if a space is composed of these joined at single points which is necessary condition possibly for any counterexample.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I think your "proof" in the planar case is too naive ("the first self-intersection" is not well-defined), and that the case of subsets of the plane would deserve a separate question (which I can't solve). $\endgroup$
    – YCor
    Commented Aug 10, 2019 at 12:11
  • $\begingroup$ What if the space embedded in $\mathbb{R}^2$ was path-connected would that help? $\endgroup$
    – Ivan Meir
    Commented Aug 10, 2019 at 12:16
  • $\begingroup$ No since the question obviously boils down to the path-connected case. $\endgroup$
    – YCor
    Commented Aug 10, 2019 at 12:22
  • 1
    $\begingroup$ This might accumulate at $x=0$. $\endgroup$
    – YCor
    Commented Aug 10, 2019 at 12:59
  • 2
    $\begingroup$ Have in mind a Peano curve. There exists $f:[0,1]\to [0,1]^2$ that is injective on no nonempty open interval (and such that the image of any nonempty open interval has nonempty interior). $\endgroup$
    – YCor
    Commented Aug 10, 2019 at 13:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .