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Assume that $X$ is a path-wise connected Hausdorff space, and assume that its fundamental group is non-trivial. Does it always exist a simple curve in $X$ which is non-null-homotopic?

Such curve does exist if we further assume that $X$ possesses a universal cover.

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  • $\begingroup$ could you provide a reference for the statement in the case that $X$ possesses a universal cover? $\endgroup$
    – Aerinmund Fagelson
    Nov 9, 2018 at 11:54
  • $\begingroup$ For future visitors, the answer is Yes if $X \subseteq \mathbb R^2$ mathoverflow.net/questions/338073/… $\endgroup$
    – D.R.
    Aug 10, 2023 at 18:57

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No. The harmonic archipelago is a compact $2$-dimensional counterexample, embedded in $\mathbb R^3$.

Let $S_n$ denote the planar circle with center on the $x$-axis, and whose intersection with the $x$-axis is $$ \left\{\Big(\frac{1}{n+1},0\Big), \Big(\frac{1}{n},0\Big)\right\}. $$ Take the closure of the union of the circles. Now make each circle the base of a cone, raised in the $z$-direction, of height $1$, and take the closure of the union of the cones. Every simple closed curve is inessential, but the space is not simply connected. (A curve running once around each base circle is essential).

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  • $\begingroup$ Your space contains the segment $\{(0,0,t) : t\in[0,1]\}$,which appears to cause some trouble. Why do you need the space to be compact? Could you provide some further explanation or a link? $\endgroup$
    – smyrlis
    Oct 14, 2015 at 15:08
  • $\begingroup$ Every simple curve is contained in one of the cones. (The example is fine and elegant, of course). $\endgroup$
    – Włodzimierz Holsztyński
    Oct 14, 2015 at 19:39
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    $\begingroup$ Cont. (finish): The z=0 level curve cannot be contracted because an infinite sequence of the level z=0 points would have to reach the top level z=1 at about the same time during the homotopy. This reaching z=1 would be forced by elementary considerations or by a simple homological argument. $\endgroup$
    – Włodzimierz Holsztyński
    Oct 14, 2015 at 19:45

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