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Let $S^{-}=\left\{0\right\}\times [-1,1]$ and $S^{+}= \left\{(x,\sin(1/x):x\in ]0,1]\right\}$ and $S=S^{-}\cup S^{+}$ (topologist's sine curve)

Hi. I would like to know if the following statement that seem intuitive to me are correct (and then try to proves them rigorously)

It is known that $S$ is connected but not path connected but: $S^{-}$ and $S^{+}$ are simply connected? i.e $S^{\pm }$ is path connected and for any $x_0\in S^{\pm}$, $\pi_{1}(S^{\pm},x_0)=\left\{[e_{x_0}]\right\}$ ($e_{x_0}$ is trivial loop in $x_0$)

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Visualization: Both $S^-$ and $S^+$ are contractible. It is easy to show that $S^-$ contracts to the point $(0,0)$. And note that $S^-$ is homeomorphic to the the open-closed interval $(0,1]$ as you can stretched and straighten the curve $S^+:=\big\{\big(x,\sin\frac{1}{x}\big) :0<x\leq 1\big\}$ to make it a line segment.

Proof: Note that $S^-$ is homeomorphic to $[-1,1]$ and $H_t:[-1,1]\ni x\longmapsto (1-t)x\in [-1,1]$ for all $0\leq t\leq 1$ defines a contraction of $[-1,1]$ to $0\in [-1,1]$.

Next note that $S^+$ is homeomorphic to $(0,1]$ via projection on the first component i.e. $S^+\ni \big(x,\sin\frac{1}{x}\big)\longmapsto x\in (0,1]$. Now, $(0,1]$ is a contactible, so we are done.

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  • $\begingroup$ I get it. So all the fundamental group in the topologist'sine curve are isomorphic and they are all trivial. $\endgroup$
    – eraldcoil
    Sep 23, 2020 at 23:54
  • $\begingroup$ The fundamental groups of $S^+$ and $S^-$ are trivial. $\endgroup$
    – Sumanta
    Sep 24, 2020 at 6:14
  • $\begingroup$ $\pi_1(S)$ is not trivial? $\endgroup$
    – eraldcoil
    Sep 25, 2020 at 4:32
  • $\begingroup$ Note that $S$ is not path-connected, and it has exactly two path-component, $S^-$ and $S^+$. So, you can't say $\pi_1(S)$, rather you have to specify a point, i.e. either you have to consider $\pi_1(S,x_0)$ where $x_0\in S^-$ and in this case $\pi_1(S,x_0)=\pi_1(S^-,x_0)=0$ or you have to consider $\pi_1(S,y_0)$ where $y_0\in S^+$ and in this case $\pi_1(S,y_0)=\pi_1(S^+,y_0)=0$. $\endgroup$
    – Sumanta
    Sep 25, 2020 at 4:41
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    $\begingroup$ Yeah, if $X'$ is the path-component of $X$ containing $x_0$. $\endgroup$
    – Sumanta
    Sep 25, 2020 at 4:48

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