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The image is a ring torus: a circle of radius $b$ is rotated about the vertical axis at distance $a$ from that axis. Put $r=\sqrt{x^2+y^2}$. Since $a>b$, every image point has $r>0$, and the image is preciselyConversely, on this level set the pairs $((r-a)/b,z/b)$ and $(x/r,y/r)$ are unit vectors, so choosing their angles gives a preimage.

The specified angular rectangle removes two closed circles from the torus. More explicitly its image isHere $C_u$ is the outer equatorial circle and $C_v$ is a meridian. The given openness property makes $U$ open in $T$. On $U$, the angles of $(r-a)+iz$ and $x+iy$ are uniquely chosen in $(0,2\pi )$; their continuous argument branches give the inverse $(x,y,z)\mapsto (u,v)$. Thus the restriction is a homeomorphism onto $U$, and the inverse is locally smooth.

For the immersion condition, the two derivatives have inner productsThey are linearly independent everywhere, since $a+b\cos u\ge a-b>0$. Hence this is a smooth embedded surface parametrization.

The angular cuts do not represent singularities of the embedded torus of revolution. To give a direct proof covering all points, define $F=(\sqrt{x^2+y^2}-a{)}^2+z^2-b^2$ on the open region $r>0$. On $T$,The implicit function theorem therefore describes $T$ near every point as a smooth graph of two coordinates. The whole torus is a smooth embedded surface.

Define a continuous map $F:S\to T$ by the standard embedded torus of revolution parametrizationIts image lies in $T$ becauseThe periodicity of the 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

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