 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/21f/a/solution

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

S^{3} = {(z_{1}, z_{2}) \in C^{2} : ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = 1} .

It is the union of the two closed solid tori

U^{'} = {∣ z_{1} ∣ \geq ∣ z_{2} ∣}, V^{'} = {∣ z_{2} ∣ \geq ∣ z_{1} ∣} .

Explicit homeomorphisms from the two copies in the question are

Φ_{U} (u, w) = \frac{( u , w )}{1 + ∣ w ∣ ^{2}}, (u, w) \in S^{1} \times D^{2},

and

Φ_{V} (z, v) = \frac{( z , v )}{1 + ∣ z ∣ ^{2}}, (z, v) \in D^{2} \times S^{1} .

On their boundaries both maps send

(u, v) \in S^{1} \times S^{1}

(u, v) / 2

. They therefore descend through the quotient to a continuous bijection

S \to S^{3}

. Since

S

is compact and

S^{3}

is Hausdorff, this bijection is a homeomorphism.

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