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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.

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  • $\begingroup$ 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
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    $\begingroup$ Recall a continuous, bijective map from a compact space to a Hausdorff space is a homeomorphism. $\endgroup$ – Joe Aug 1 '12 at 14:04
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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 $D^2$.

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