Exercise 1.1.16

Proof of $(a)$. Since ${ℝ}^3$ is convex, ${\pi }_1(X)=0$ by Example 1.4. Also, ${\pi }_1(A)=\mathbb{Z}$ by Theorem $1.7$. But $\mathbb{Z}$ does not inject into $0$, contradicting Proposition $1.17$. □

Proof of $(b)$. Since $X$ deformation retracts to its center circle, ${\pi }_1(X)=\mathbb{Z}$ by Proposition $1.17$. However, ${\pi }_1(A)={\mathbb{Z}}^2$ by Example $1.13$. But ${\mathbb{Z}}^2$ does not inject into $\mathbb{Z}$, for every non-trivial subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, and free $\mathbb{Z}$-modules of different rank are non-isomorphic, contradicting Proposition $1.17$. □

Proof of $(c)$. Recall that $X$ deformation retracts to its center circle $C$, and so we have the chain $A\stackrel{i}{\text{↪}}X\stackrel{p}{\to }C$ which induces the maps ${\pi }_1(A)\stackrel{i_{\text{∗}}}{\to }{\pi }_1(X)\stackrel{p_{\text{∗}}}{\to }{\pi }_1(C)$, where $p_{\text{∗}}$ is an isomorphism by Proposition $1.17$. But if $f$ is a representative of the generator of ${\pi }_1(A)=\mathbb{Z}$, we see that $(i\circ p)(f)$ is nulhomotopic since it has the same start and end points when lifted up into $ℝ$ as in Theorem $1.7$, and so $(i_{\text{∗}}\circ p_{\text{∗}})(1)=0$ the constant loop. Hence, $i_{\text{∗}}$ is not an injection, contradicting Proposition $1.17$. □

Proof of $(d)$. By Example $1.21$, ${\pi }_1(X)={\pi }_1(D^2)\ast {\pi }_1(D^2)=0$ by Example $1.4$ since $D^2$ is convex. However, ${\pi }_1(A)=\mathbb{Z}\ast \mathbb{Z}$ again by Example $1.21$. $\mathbb{Z}\ast \mathbb{Z}$ does not inject into $0$, contradicting Proposition $1.17$. □

Proof of $(e)$. We see that $D^2$ with a $1$-cell attached to two points on the boundary is homotopy equivalent to $X$, since the $1$-cell is contractible. On the other hand, since $D^2$ is contractible, $X$ is homotopy equivalent to $S^1$. Thus, ${\pi }_1(X)=\mathbb{Z}$, while ${\pi }_1(A)=\mathbb{Z}\ast \mathbb{Z}$ by Example $1.21$. Since $\mathbb{Z}\ast \mathbb{Z}$ is not abelian while all subgroups of $\mathbb{Z}$ are abelian, $\mathbb{Z}\ast \mathbb{Z}$ does not inject into $\mathbb{Z}$, contradicting Proposition $1.17$. □

Proof of $(f)$. Recall that $X$ deformation retracts to its center circle $C$, and so we have the chain $A\stackrel{i}{\text{↪}}X\stackrel{p}{\to }C$ which induces the maps ${\pi }_1(A)\stackrel{i_{\text{∗}}}{\to }{\pi }_1(X)\stackrel{p_{\text{∗}}}{\to }{\pi }_1(C)$, where $p_{\text{∗}}$ is an isomorphism by Proposition $1.17$. But if $f$ is a representative of a generator of ${\pi }_1(A)=\mathbb{Z}$, we see that $(i\circ p)(f)$ maps to $g^2$ where $g$ is a representative of the generator of ${\pi }_1(C)=\mathbb{Z}$, and so $(i_{\text{∗}}\circ p_{\text{∗}})(1)=2$, and in particular $i_{\text{∗}}(1)=2$. This contradicts that $i_{\text{∗}}\circ r_{\text{∗}}={\mathrm{id}<!--\; FUNCTION\; APPLICATION\; -->}_{\pi (A)}$ since every homomorphism $\mathbb{Z}\to \mathbb{Z}$ is of the form $n\mapsto \mathit{kn}$ for some $k\in \mathbb{Z}$, and so there is no homomorphism such that $2\mapsto 1$. □