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I was trying to show that fundamental group of quasi circle(page 79 ,Hatcher) is trivial.I can understand that every loop is precisely zero loop because for any loop if it start at some point it has to come back at that point and there is exactly one path so it has to follow the same path to return so this is just composition of path with its inverse (some kind of). so that will give zero loop.but i want if there is any other precise way to prove this fact? Thank you in advance

Shivani Sengupta
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Let $Y$ be the Warsaw circle, i.e., the graph of $y = \sin(1/x)$ for $x \in (0,1]$, along with the segment $[-1,1]$ in the $y$-axis, and with an arc connecting both pieces. Since the image of any loop in $Y$ is compact, it must belong in $X = Y \setminus \{(x,\sin(1/x)) \mid x < a\}$ for some $a > 0$. (Consider a family of open covers of the form $((1/n,1) \times (-2,2)) \cap Y$ as $n$ goes to infinity, along with an open set which intersects a finite number of members of that family and which contains part of the arc. If necessary, contract any part of the loop that lies inside the $y$-axis into the arc to ensure that the cover contains the whole image of the loop.) But $X$ is contractible; therefore any loop in $Y$ can be contracted to a point.

Alex Provost
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    I doubt the existence of such open set "which intersects a finite number of members of that family and which contains part of that arc". This open set contains the point $(0,0)$ especially. However, for any open set containing $(0,0)$, it must intersect infinitely many $(1/n,1)\times(-2,2)$. – Andrews Jan 19 '20 at 13:07
  • Indeed, the Warsaw circle is compact by something like the argument in @Andrews' comment. Concretely, any open cover must cover the compact $[-1,1]$ segment, so we may extract a finite subcover of the $[-1,1]$ segment. This subcover will contain an $\epsilon$ collar of the segment, so it covers the pathological bit of the circle. – Zach Effman Sep 01 '20 at 01:52