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Given a topological space $X$ and a path $p : [0, 1] \to X$ such that $p(0) \neq p(1)$, does there always exist an injective path $p' : [0,1] \to \operatorname{im}p$ such that $p'(0) = p(1)$ and $p'(1) = p(1)$? For intuitive examples of paths we could "cut out" the loops which occur in $p$, but are there pathological examples of paths in which this technique cannot be applied? Otherwise, is there a Theorem that guarantees the existence of $p'$, even for only certain "nice" classes of paths $p$?

For any $X$ that satisfies this property, any subspace of $X$ is path-connected if and only if every two points are joined by an injective path.

What about the special case where $X = \mathbb{R}^n$?

Herng Yi
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  • You want to assume $p(0) \neq p(1)$ to ensure injectivity is possible, i.e. that $p$ is not a loop. It seems to me you might be able to do some sort of inductive process where you cut out loops, but it's going to get messy if $p$ is a space-filling curve. – JSchlather Feb 05 '13 at 03:29
  • Thanks, I'll add that into the question. – Herng Yi Feb 05 '13 at 03:35
  • I may misunderstand the question, but I think that it is probably not true in general because there are countable, connected Hausdorff spaces. In particular, since $[0,1]$ is uncountable, there cannot be an injection into a countable space, and we require it be connected since the continuous image of a connected set is connected. – Clayton Feb 05 '13 at 04:08
  • @Clayton Do you have an example of a non-trivial path-connected countable space $X$? – JSchlather Feb 05 '13 at 04:52
  • @JacobSchlather: A quick search returned this page, but I haven't taken the time to read it, so it might not work. – Clayton Feb 05 '13 at 04:54
  • @Clayton, Yes I'm not sure about that one but spacebook says this space is path-connected. So perhaps OP should ask the question in a nicer setting, such as Euclidean space or manifolds. – JSchlather Feb 05 '13 at 05:01
  • @JacobSchlather: Yes, I am prepared to ask the question in the setting of Euclidean space. I will edit the question to reflect that. – Herng Yi Feb 05 '13 at 12:33
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    This seems to answer you question. – Miha Habič Feb 05 '13 at 13:14
  • @MihaHabič It does... I'm unhappy though, because the intuition of "cutting out loops" suggests we should have a simple loop of the form $p'=p\circ f$ where $f:[0,1]\to[0,1]$ is continuous, nondecreasing and fixes the endpoints. An argument based on Hahn-Mazurkiewicz does not give this, as far as I can tell. Perhaps the question can be reworded again to ask for $p'$ of the above kind... better than just closing it as a duplicate. –  Feb 05 '13 at 16:12
  • @5PM I'm not sure I understand what you were trying to say with $p'=p\circ f$; if $f$ is as you require, it is surjective so $p'$ would be surjective onto the image of $p$. – Miha Habič Feb 05 '13 at 18:51
  • @MihaHabič Sorry, $f$ should not be continuous. It should be nondecreasing, fixing the endpoints, and such that $p'=p\circ f$ is a continuous simple path. The jumps of $f$ correspond to loops of $p$ being cut out. –  Feb 05 '13 at 18:55
  • @5PM Maybe something like $$f(t)=\max{\sup p^{-1}[{t}],\sup_{s\lt t}f(s)}$$ would work? Possibly with some padding to fix the endpoints? – Miha Habič Feb 05 '13 at 19:24

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