Showing or refuting that topologist's sine curve is simply connected.

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 $$x_0$$)

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 $$S^+:=\big\{\big(x,\sin\frac{1}{x}\big) :0 to make it a line segment.
Proof: Note that $$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.
• The fundamental groups of $S^+$ and $S^-$ are trivial. Sep 24, 2020 at 6:14
• $\pi_1(S)$ is not trivial? Sep 25, 2020 at 4:32
• 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 $\pi_1(S,y_0)=\pi_1(S^+,y_0)=0$. Sep 25, 2020 at 4:41
• Yeah, if $X'$ is the path-component of $X$ containing $x_0$. Sep 25, 2020 at 4:48