Prove that the cone over $S^1$ is homeomorphic to $D^2$ 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.
 A: 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 $D^2$.
