Exercise 22.3

Proof. An illustration of the subspace $A\subset ℝ\times ℝ$ is shown below:

First, we know that ${\pi }_1$ is continuous from §18. It then follows that the restriction $q$ is a continuous map as well by Theorem 18.2 part (d).

Now define a map $f:ℝ\to A$ by $f(x)=x\times 0$ for any $x\in ℝ$, noting that clearly $f(x)\in A$. We show that $f$ is continuous by considering any basis element $U\times V$ of $ℝ\times ℝ$ so that $U$ and $V$ are open in $ℝ$. Now, if either of $U$ or $V$ are empty, then of course $U\times V$ is empty as well so that $f^{-1}(U\times V)=f^{-1}(\varnothing )=\varnothing$ is open in $ℝ$. Otherwise if $0\notin V$ then also $f^{-1}(U\times V)=\varnothing$ is open in $ℝ$. If $0\in V$ then we claim that $f^{-1}(U\times V)=U$, which is of course open in $ℝ$. This is easy to show:

since we know that $0\in V$. Thus in all cases $f^{-1}(U\times V)$ is open in $ℝ$, which suffices to show that $f$ is continuous.

Now consider any $x\in ℝ$ so that we have

which shows that $q\circ f=i_{ℝ}$. Since $q:A\to ℝ$ and $f:ℝ\to A$ have both been shown to be continuous, it follows from Exercise 22.1 part (a) that $q$ is a quotient map as desired.

To show that $q$ is not an open map, consider that subset $U=[0,1)\times (1,2)\subset A$, which is open in the subspace $A$ since $U=A\cap [(-1,1)\times (1,2)]$ and clearly $(-1,1)\times (1,2)$ is a basis element of $ℝ\times ℝ$ and so is open. However, clearly the set $q(U)={\pi }_1(U)=[0,1)$ is not open in $ℝ$. To show that $q$ is not a closed map, consider the set $C=\left\{x\times (1∕x)\mid x>0\right\}$. It is easy to see and not difficult to show that $C$ is a closed subset of the subspace $A$ because no point of $A-C$ is a limit point of $C$. Also clearly $q(C)={\pi }_1(C)={ℝ}_{\text{+}}$, which is not closed in $ℝ$ since its complement $\left\{x\mid x\le 0\right\}$ is not open. Thus $q$ is not a closed map either. □