13
$\begingroup$

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.

enter image description here

$Y$ is simply-connected: Let $$A=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):0<x\leq 1\bigg\}\text{ and }X=\big\{(0,y):-1\leq y\leq 1\big\}\bigcup A.$$ 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 $$A_\delta=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):\delta<x\leq 1\bigg\}\text{ and }X_\delta=A_\delta\bigcup\big\{(0,y):-1\leq y\leq 1\big\} ,$$ $$Y_\delta:=X_\delta\cup C.$$ 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 $$\text{Image}(\alpha)\cap B_{n}\not=\emptyset\text{ where }B_n:=\bigg\{\bigg(x,\sin \frac{1}{x}\bigg):0<x<\frac{1}{n}\bigg\}.$$ 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 $t_m<t_k^*<t_n$ 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 $\big\{x\in\Bbb R:\alpha_2(c)-\varepsilon<x<\alpha_2(c)+\varepsilon\big\}$, where $\varepsilon>0$ is so chosen that $\big\{x\in\Bbb R:\alpha_2(c)-\varepsilon<x<\alpha_2(c)+\varepsilon\big\}$ 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.

enter image description here

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 $\big\{\big(x,\sin \frac{1}{x}\big):0<x\leq 1\big\}\subseteq X$ 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 $p^{-1}\big([z]\big)\cap \{x\in\Bbb R:0<x<1\}$ is singleton.

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

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

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

$\endgroup$
1
  • 1
    $\begingroup$ Your space $Y$ is known as the Warsaw circle. The (closed) topologist's sine curve is the space $X$. $\endgroup$
    – Paul Frost
    Oct 27, 2022 at 23:48

0

You must log in to answer this question.