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If $\gamma:S^1\to X$ is a simple contractible loop, when can we say there must by a contraction, $H(s,t)$ such that $\gamma_t:s\mapsto H(s,t)$ is a simple loop for all $t<1?$

(1) It seems like you should be able to do this if $X$ is a manifold. (Of dimension $>1,$ but there are no retractable simple loops in manifolds of dimension $1.$) Maybe it requires dimension $>2?$

(2) I’d like to find a Hausdorff counterexample, or a proof that we can always find $H$ in those cases.

When there is a counterexample, the base case would be a quotient of the closed disk, with no more than one point on the boundary in each equivalence class. With $\sim$ defined cleverly enough, then $D^2/\sim$ with the loop $S^1\to D^2\to D^2/\sim $ would be an example.

A simple non-Hausdorff counterexample is $u\sim v$ if both $|u|,|v|<1/2.$ This quotient is non-Hausdorff, since an equivalence class is not closed. The singleton in $D^2/\sim$ consisting of the equivalence class of $0$ is open in $D^2/\sim.$ So you can’t cross $[0]$ to retract to another point - $\gamma_{t}^{-1}([0])$ must be an open set. And you can’t retract to $[0]$ point since $H^{-1}(\{[0]\})$ would have to be open, so must contain an $(s,t)$ with $t<1.$

If $X$ and $\gamma$ are a counterexample, then this means in particular that there is a $P:D^2\to X$with $P_{|S^1}=\gamma.$

If $X$ is Hausdorff, then when we define $u \sim_P v$ iff $P(u)=P(v),$ then $Y=D^1/\sim_P$ is homeomorphic to $P(D^2),$ since $Y$ is compact and $P(D^2)$ is Hausdorff.

If we can’t find a proper retraction in all of $X,$ we can’t find one in $P(D^2)$ and hence not in the case of $\gamma’:S^1\to D^2/\sim_P.$

So any Hausdorff counterexample gives us a Hausdorff counterexample of the form $D^2/\sim$ for some equivalence relation $\sim.$

Thomas Andrews
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  • Rectractable = contractible? – Paul Frost Sep 21 '21 at 22:51
  • Yeah, I forgot the distinction,. Edited. @PaulFrost – Thomas Andrews Sep 21 '21 at 22:57
  • Hmm, my last section about $D^1/\sim$ seems wrong. At least, with example in the answer below, we end up with $\sim$ identifying just two points in the interior, and it seems like we can do the homotopy there by hitting that point twice in different times. – Thomas Andrews Sep 22 '21 at 03:37
  • @ThomasAndrews: It's not just two points in the interior that get identified in that example. As you pull the two ends of the top part of the knot apart, they will cross over the same region in space. There is only one point where they overlap at the same time, but there are many points that get hit by both sides at different times. – Eric Wofsey Sep 22 '21 at 03:51
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    Also, I think everywhere you wrote $D^1$ you mean $D^2$ instead. – Eric Wofsey Sep 22 '21 at 03:52
  • Oops, yes, fixed. Thanks. @EricWofsey – Thomas Andrews Sep 22 '21 at 04:39

1 Answers1

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Here's one manifold counterexample in three dimensions:

Let $X=\left(\mathbb{R}^2\setminus\{0\}^2\right)\times\mathbb{R}$ be $\mathbb{R}^3$ sans the $z$-axis, and consider the embedding $S^1\to X$ given in the following "knot diagram":

enter image description here

Where the $z$-dimension is suppressed. The loop is clearly contractible, but it cannot be contracted in the desired manner. To see this, suppose such an $H$ exists, and note that $\gamma_t$ is an unknot in $\mathbb{R}^3$ for all $t<1$, and there is an $\epsilon>0$ such that $\gamma_{1-\epsilon}$ is contained in an open ball disjoint from the $z$-axis. Adding a point at infinity, $H|_{S^1\times[0,1-\epsilon]}$ then becomes an isotopy which unlinks the Whitehead link, which is a contradition.

Kajelad
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