 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/21h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

g : T \to T, g (u, v) = (u v^{- 1}, v),

is continuous and satisfies

g \circ f = f \circ g = id_{T}

. Therefore

f^{- 1} (u, v) = (u v^{- 1}, v), so f is a homeomorphism .

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