Exercise 71.5

Proof of $(a)$. $Y$ is a subspace of ${ℝ}^2$ and so it is first countable by Theorem $30.2$, and so $Y$ is not homeomorphic to $X$ by Exercise , which says $X$ is not first countable.

Denote $Z$ as the space of Example $71.1$; we claim $Z$ is compact while $Y$ is not. First, denoting $D_m$ as the closed disc of radius $1∕m$ with center at the point $(1∕m,0)$, we see that $Z\cup D_m$ is closed since it is a finite union of closed sets $Z\cup D_m=D_m\cup {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n<m}{C}_n$. Then, $Z\subseteq {\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_m(Z\cup D_m)$ trivially; moreover, since if $x\notin Z$, then $x\notin D_m$ for all $m$ large enough, we have that $Z={\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_m(Z\cup D_m)$. Thus, $Z$ is closed. Since $Z$ is bounded, it is then compact by Theorem $27.3$. On the other hand, $Y$ is unbounded hence not compact by Theorem $27.3$, and so $Y,Z$ are not homeomorphic. □

Proof of $(b)$. $\ast \ast$ Let $i_n:{\pi }_1(S_n,p)\to {\pi }_1(Y,p)$ be the homomorphism induced by inclusion, and let their respective images be $G_n$; we want to show that the homomorphism

is an isomorphism.

We first show that the homomorphism (1) is surjective. So, suppose $f:I\to Y$ is a loop in $Y$. Letting $Y_N=Y\setminus {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=N+1}^{\infty }\{(2n,0)\}$, since $I$ is compact, the image $f(I)$ is also compact, thus bounded by Theorem $27.3$, and so $f(I)\subseteq Y_N$ for some $N$. Letting $r_N:Y\to {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{S}_n$, we form the deformation retraction $H:Y_N\times I\to Y_N$, where $H(x,0)={\mathrm{id}<!--\; FUNCTION\; APPLICATION\; -->}_{Y_N}$ and $H(x,1)=r_N{\text{|}}_{Y_N}$, by retracting the upper and lower semicircles of $S_n$ to $p$ for all $n>N$. Then, $H\circ (f\times {\mathrm{id}<!--\; FUNCTION\; APPLICATION\; -->}_I):I\times I\to Y_N$ is a path homotopy from $f$ to $r_N\circ f$. Thus, $[f]=[r_N\circ f]$, and Theorem $71.1$ implies that $[r_N\circ f]$ is a product of elements of the groups $G_n$ for $n\le N$. Therefore, $[f]$ is in the image of the homomorphism (1).

We now show that the homomorphism (1) is injective. Now suppose there is some non-identity element

whose image through the homomorphism (1) is the identity in ${\pi }_1(Y,p)$. Let $f$ be a loop in $X$ whose path-homotopy class is the image of $w$. Then, $f$ is path homotopic to a constant in $X$, so by the argument above, it is path homotopic to a constant in some $Y_N$, and therefore path homotopic to a constant in ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^N{S}_n$. But this contradicts Theorem $71.1$, which says that the map

is injective. □