Let \(X=\RR\times\RR\) with the product topology \(\mathcal T\), giving \(\RR\) the usual metric topology, and let \(p:(X,\mathcal T)\to(\RR,\text{usual})\) be the projection on the first coordinate, \(p(x,y)=x\). Let \(A=\{(x,y)\in X:x\ge0\text{ or }y=0\}\), with the subspace topology that it inherits from \(X\). Show that \(f=p|_A:A\to\RR\) is a quotient map, but that \(f\) is neither an open map nor a closed map.
Solution
1. \(f\) is continuous and surjective.
\(f\) is a restriction of the continuous projection \(p\), and \((x,0)\in A\) maps to \(x\) for every \(x\in\RR\).
2. If \(V\subseteq\RR\) and \(f^{-1}(V)\) is open in \(A\), then \(V\) is open in \(\RR\).
Fix \(x\in V\). Since \((x,0)\in f^{-1}(V)\), there is an open rectangle \(R=(x-\varepsilon,x+\varepsilon)\times(-\eta,\eta)\) with \(R\cap A\subseteq f^{-1}(V)\). For every \(t\in(x-\varepsilon,x+\varepsilon)\), \((t,0)\in R\cap A\), so \(t=f(t,0)\in V\).
3. \(f\) is not open.
\(O=A\cap(\RR\times(1,2))=[0,\infty)\times(1,2)\) is open in \(A\), but \(f(O)=[0,\infty)\) is not open in \(\RR\).
4. \(f\) is not closed.
\(C=\{(x,y)\in\RR^2:xy=1\}\) is closed in \(\RR^2\) as the preimage of \(\{1\}\) under \((x,y)\mapsto xy\). No point of \(C\) has \(y=0\), so \(C\cap A=\{(x,1/x):x>0\}\), which is closed in \(A\). But \(f(C\cap A)=(0,\infty)\) is not closed in \(\RR\).
5. Q.E.D.
By steps 1 and 2, \(f\) is a quotient map, and by steps 3 and 4 it is neither open nor closed.