Let \(\pi_1: \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) be projection on the first coordinate. Let \(A\) be the subspace of \(\mathbb{R} \times \mathbb{R}\) consisting of all points \(x \times y\) for which either \(x \geq 0\) or \(y = 0\) (or both); let \(q: A \to \mathbb{R}\) be obtained by restricting \(\pi_1\). Show that \(q\) is a quotient map that is neither open nor closed.
Solution
1. \(q\) is continuous and surjective.
\(q\) is the restriction of the continuous projection \(\pi_1\), and \(q(x,0)=x\) with \((x,0)\in A\) for every \(x\in\mathbb R\).
2. \(q\) is a quotient map.
By step 1 it suffices to show that \(U\subseteq\mathbb R\) is open whenever \(q^{-1}(U)\) is open in \(A\). Let \(x_0\in U\). Then \((x_0,0)\in q^{-1}(U)\), so there is an open box \(B=(a,b)\times(-\varepsilon,\varepsilon)\) containing \((x_0,0)\) with \(B\cap A\subseteq q^{-1}(U)\). For \(t\in(a,b)\), the point \((t,0)\) lies in \(B\cap A\), so \(t=q(t,0)\in U\). Hence \(x_0\in(a,b)\subseteq U\).
3. \(q\) is not open.
The set \([0,\infty)\times(0,\infty)=A\cap(\mathbb R\times(0,\infty))\) is open in \(A\), and its image \([0,\infty)\) is not open in \(\mathbb R\).
4. \(q\) is not closed.
The set \(C=\{(x,y):x>0,\ xy=1\}\) equals \(\{(x,y):x\ge0,\ xy=1\}\), which is closed in \(\mathbb R^2\); it lies in \(A\), so it is closed in \(A\). Its image \(q(C)=(0,\infty)\) is not closed in \(\mathbb R\), since \(0\) is a limit point of it.
5. Q.E.D.
Steps 2, 3, and 4.