# Topologist's sine curve is a simply-connected space

I am trying to solve the following problem from Hatcher's Algebraic Topology and have written a solution. Could you help me checking my solution, whether I am right? Thanks in advance.

$$Y$$ is simply-connected: Let $$A=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):0 and $$C$$ be a simple curve in $$\Bbb R^2$$ such that $$C\cap X=\big\{(0,0),\big(1,\sin 1\big)\big\}$$. Define $$Y=X\cup C$$. Now, for each $$1\geq \delta>0$$ let $$A_\delta=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):\delta $$Y_\delta:=X_\delta\cup C.$$ Note that $$Y=\bigcup_{0<\delta\leq 1}Y_\delta.$$ Note that $$Y$$ is path-connected, we show $$\pi_1\big(Y,(0,0)\big)=0$$. Let $$\alpha=(\alpha_1,\alpha_2)$$ be a loop in $$Y$$ based at $$(0,0)$$. If possible let for each $$n\in \Bbb N$$ we have $$\text{Image}(\alpha)\cap B_{n}\not=\emptyset\text{ where }B_n:=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):0 Take, $$\alpha(t_n)=(x_n,y_n)\in \text{Image}(\alpha)\cap B_{n}$$ for each $$n\in\Bbb N$$, so $$\alpha_1(t_n)=x_n\to 0$$. Since $$[0,1]$$ is compact passing to the subsequence we may assume $$t_n\to c\in [0,1]$$. So, $$\alpha_1(c)=0$$. Also, $$-1\leq \alpha_2(c)\leq 1$$ as $$X$$ is closed. Since, $$\alpha_1(t_n)=x_n\to 0$$ by intermediate value theorem $$\alpha_1$$ takes all values $$\frac{1}{n\pi}$$ for all large $$n\in\Bbb N$$. Suppose $$\frac{1}{k\pi}\in \text{Image}(\alpha_1)$$ for all $$k\in\Bbb N$$ with $$k\geq k_0$$, where $$k_0\in\Bbb N$$. Let $$\alpha_1(t_m)=x_m<\frac{1}{k\pi}<\alpha_1(t_n)=x_n$$ for some $$m,n\in\Bbb N$$. So, by intermediate value theorem we have $$t_m such that $$\alpha_1(t_k^*)=\frac{1}{k\pi}$$. Hence, $$t_k^*\to c$$ as $$k\to \infty$$. Now, $$\alpha_2(t_k^*)\in\{\pm 1\}$$ for all $$k\geq k_0$$. So, there is no neighbourhood of $$c$$ mapped into $$\big\{x\in\Bbb R:\alpha_2(c)-\varepsilon, where $$\varepsilon>0$$ is so chosen that $$\big\{x\in\Bbb R:\alpha_2(c)-\varepsilon contains exactly one element from $$\{\pm 1\}$$, as each nbd of $$c$$ contains some $$t_k^*$$. And this is a contradiction to the continuity of $$\alpha_2$$.

So, for large $$n$$ we have $$\text{Image}(\alpha)\cap B_n=\emptyset.$$

That's, $$\text{Image}(\alpha)\subseteq Y_{\delta}$$ for some $$\delta>0$$. Now, the space $$Y_{\delta}$$ is homeomorphic to the space $$\big\{(x,0):-1< x\leq 0\big\}\cup\big\{(0,y):-1\leq y\leq 1\big\}.$$ So, $$Y_{\delta}$$ is contractible. Hence, $$\pi_1\big(Y_{\delta},(0,0)\big)=0$$. In other words, there is a homotopy $$H:[0,1]\times [0,1]\to Y_{\delta}$$ such that $$H:\alpha\simeq_{\text{rel }(0,0)}C_{(0,0)}$$, So, extending the co-domain we have a homotopy $$H:[0,1]\times [0,1]\to Y$$ such that $$H:\alpha\simeq_{\text{rel }(0,0)}C_{(0,0)}$$. So, $$\pi_1\big(Y,(0,0)\big)=0$$.

Now, $$\Bbb S^1\cong\frac{C}{\big\{(0,0),(1,\sin 1)\big\}}\cong\frac{Y}{X}$$ considering inclusion map $$i:C\hookrightarrow Y$$. So, we have a quotient map $$f:Y\to \frac{Y}{X}=\Bbb S^1$$. Let $$p:\Bbb R\ni t\longmapsto e^{2\pi it}\in \Bbb S^1$$ be the universal cover.

No Lifting: Now, suppose we have continuous lifting as in the picture below.

Now, $$p^{-1}(1)=p^{-1}\big([X]\big)$$ is a discrete set, so continuity of $$\widetilde f$$ implies $$\widetilde f(0,y)=0$$ for $$-1\leq y\leq 1$$ as $$f(0,y)=[X]=1$$ for $$-1\leq y\leq 1$$.

Similarly, $$\widetilde f$$ is identically $$0$$ on $$\big\{\big(x,\sin \frac{1}{x}\big):0 using continuity.

Note that $$f$$ is surjective, so image of $$\widetilde f$$ contains either of the two intervals $$(-1,0)\subseteq \Bbb R$$ or $$(0,1)\subseteq \Bbb R$$. Without loss of generality assume, the latter one.

Next, for each point $$z\in C\backslash \big\{(0,0),(1,\sin 1)\big\}$$ the equivalence class $$[z]\in \frac{Y}{X}$$ is a singleton set and $$p^{-1}\big([z]\big)\cap \{x\in\Bbb R:0 is singleton.

So, if $$C \backslash \big\{(0,0),(1,\sin 1)\big\}\ni z\longrightarrow (0,0)$$, then $$\widetilde f(z)=p^{-1}\big([z]\big)\cap \{x\in\Bbb R:0

and if $$C \backslash \big\{(0,0),(1,\sin 1)\big\}\ni z\longrightarrow (1,\sin 1)$$, then $$\widetilde f(z)=p^{-1}\big([z]\big)\cap \{x\in\Bbb R:0,

a contradiction as $$\widetilde f$$ is identically $$0$$ on $$\big\{\big(x,\sin \frac{1}{x}\big):0.

• Your space $Y$ is known as the Warsaw circle. The (closed) topologist's sine curve is the space $X$. Oct 27, 2022 at 23:48