As the title, I am thinking $$f :\ S^1\times I / S^1\times \lbrace1\rbrace \rightarrow D^2$$ as $f(x,y,z)= (x,y)$ and $$g :\ D^2 \rightarrow S^1\times I / S^1\times\lbrace1\rbrace$$ as $g(x,y)=(x,y,1-\sqrt{x^2+y^2})$.

Just remains to show $f$ and $g$ are continuous.

  • $\begingroup$ Try mapping $(x,t)$ to $tx$ in $\mathbb{R}^2$. $\endgroup$ – user641 Aug 1 '12 at 11:47
  • $\begingroup$ Use the center of a circle in the plane as the cone point. The radii from the center gives the cone on $S^1$, so it is a $D^2$ $\endgroup$ – i. m. soloveichik Aug 1 '12 at 13:44
  • 1
    $\begingroup$ Recall a continuous, bijective map from a compact space to a Hausdorff space is a homeomorphism. $\endgroup$ – Joe Aug 1 '12 at 14:04

There is no need of such complicated formulas. Consider the map

$$f: S^1 \times I /S^{1} \times \{1\} \rightarrow D^2$$

defined by $f(\theta,t) = (1-t)e^{i \theta}$. Then $f$ is a continuous function because it is the product of two continuous functions. Now it is clear that $f$ is surjective once we write down a complex number in $D^2$ in polar form.

We now check that it is injective. If $t_1,t_2$ are not equal to 1 then whenever $t_1 \neq t_2$, for all $\theta,\phi$ we have $f(\theta,t_1) \neq f(\phi,t_2)$. This comes from the fact that the two points will have different radii. Now if $t=1$, $f$ reduces to the zero map. But this does not matter because we have quotiened out by $S^1 \times \{1\}$. Hence $f$ is bijective and is continuous with respect to the Euclidean topology on $D^2$ and the quotient topology on the quotient space. Now $S^{1} \times I$ is compact, and so $S^{1} \times I/ S^1 \times \{1\}$ is compact as well.

We have now satisfied all the hypotheses of Matt's comment from which it follows that $f$ is the required homeomorphism between the cone $CS^1$ over $S^1$ and $D^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.