Show there is a continuous surjective map \(f: \mathbb{R} \to \mathbb{R}^n\).
Solution
By E-SADP0, every compact cube \([-m,m]^n\), which is homeomorphic to \(I^n\), is a continuous image of \(I\). We may choose a continuous surjection \[ q_m:I\to[-m,m]^n \] and modify its parametrization so that it starts and ends at the origin: use the first third of \(I\) for a straight path from \(0\) to \(q_m(0)\), the middle third to traverse \(q_m\), and the last third for a straight path from \(q_m(1)\) back to \(0\). The resulting map, still denoted \(q_m\), is surjective and satisfies \[ q_m(0)=q_m(1)=0. \]
Define \(F:[0,\infty)\to\mathbb R^n\) by setting \(F(0)=0\) and, for each integer \(m\ge1\), \[ F(t)=q_m(t-m+1)\qquad(m-1\le t\le m). \] The definitions agree at integer endpoints, so the pasting lemma gives continuity. Since the cubes \([-m,m]^n\) exhaust \(\mathbb R^n\), \(F\) is surjective. Finally define \[ f(t)=F(|t|),\qquad t\in\mathbb R. \] Then \(f:\mathbb R\to\mathbb R^n\) is continuous and surjective.