Bourbaki

Exercise 3, § 2

General Topology, VIII

Show, without using Theorem 1, that the mapping z→z/z‾z \to z/\overline{z} is a strict morphism of the topological group C∗\mathbf{C}^* onto the topological group U\mathbf{U}. Deduce that the mapping t→(1+it)/(1−it)t \to (1 + it)/(1 - it) [(1+it)/(1−it)=−1(1 + it)/(1 - it) = -1 if t=∞t = \infty] is an isomorphism of the topological group R~\tilde{\mathbf{R}} (no. 6) onto the topological group U\mathbf{U}, and hence give another proof of Theorem 1.

Solution

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Transcribed from the printed volume. Not translated.