Let $\Phi :[0,2\pi] \times [0,2\pi] \rightarrow R^{3}$ be given by $$\Phi(\theta,\phi)=((A+\cos\phi)\cos\theta,(A+\cos\phi)\sin\theta,\sin\phi), \quad A>1 $$ Explicitly prove that image of $\Phi$ is homeomorphic to $S^{1}\times S^{1}$.

First of all,$[0,2\pi]$ is homeomorphic to $S^{1}$ by gluing 0 and $2\pi$ together via exponential function. Then by introducing an equivalent relationship on $[0,2\pi] \times [0,2\pi]$,we have a quotient map.from university property of quotient map,we know that map induced from $\Phi$ which is from $S^{1}\times S^{1}$ to image of $\Phi$ is continuous .Surjective is obvious.I don't know how to prove injective and continuity of the inverse. Could anyone give some help?


1 Answer 1


This is an application of a standard theorem of topology which you can find, for example, in Munkres "Topology", which goes like this.

If $f : X \to Y$ is a continuous, surjective function from a compact space $X$ to a Hausdorff space $Y$, then $f$ is a quotient map (the proof is elementary, and uses the fact that a compact subset of a Hausdorff space is closed, and a closed subset of a compact Hausdorff space is compact).

Furthermore, let $Q$ be the quotient space of $X$ obtained by decomposing $X$ into the point pre-images $f^{-1}(y)$, $y \in Y$. Also let $F : Q \to Y$ be the induced function $F(f^{-1}(y))=y$. Then $F$ is a homeomorphism (again the proof is elementary, being just an application of the definition of a quotient map and the quotient topology).

In your example, $X = [0,1] \times [0,1]$ which is clearly compact, $Y$ is a subspace of $\mathbb{R}^3$ which is clearly Hausdorff, and $f$ is clearly continuous and surjective.


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