= Solution
Define a continuous map $F:S\to T$ by the standard <embedded torus of revolution> parametrization
$$
F(s,t)=\bigl((2+\cos2\pi t)\cos2\pi s,
(2+\cos2\pi t)\sin2\pi s,
\sin2\pi t\bigr).
$$
Its image lies in $T$ because
$$
\sqrt{x^2+y^2}=2+\cos2\pi t,
\qquad
(\sqrt{x^2+y^2}-2)^2+z^2=1.
$$
The periodicity of the <trigonometric function>[trigonometric functions] makes $F(s,0)=F(s,1)$ and $F(0,t)=F(1,t)$, so $F$ is constant on every $\sim$-equivalence class. The <universal property of the quotient topology> therefore gives a unique continuous map
$$
\overline F:S/{\sim}\longrightarrow T,
\qquad
\overline F([s,t])=F(s,t).
$$
Every point of $T$ has cylindrical radius between $1$ and $3$, so its polar angle determines $s$ modulo one, while the pair $(\sqrt{x^2+y^2}-2,z)$ on the <unit circle> determines $t$ modulo one. Hence $\overline F$ is surjective, and two parameters have the same image exactly when they differ by the endpoint identifications defining $\sim$; thus it is injective.
The quotient $S/{\sim}$ is compact as the continuous image of the given compact space $S$ under the <quotient map>. The surface $T$ is Hausdorff because it has the <subspace topology> inherited from $\mathbb R^3$. The <compact-to-Hausdorff continuous bijection theorem> now shows that $\overline F$ is a homeomorphism. Therefore
$$
\boxed{S/{\sim}\cong T}.
$$
Back to article page
