Cone over a topological space: building it. I have a couple of questions involving cones over topological spaces and their homeomorphisms with the closed disc $D^2$.
First of all, let's build a cone over a topological space. For example, let's consider the unit circle $S^1$ on $\mathbb{R}^2$. As I understood it, we firs start with a cylinder $$S^1 \times [0,1].$$
Now, we have to collapse one "end" of the cylinder into a point. This point will be the identity of the space $\mathbb{R}^2$. So let's consider
$$\ (S^1\times [0,1])\  /\  (S^1\times \lbrace1\rbrace).$$
We have a cone with the "base" on the floor, and the collapsed point on the top of the cone.
If we consider $$\ (S^1\times [0,1])\  /\  (S^1\times \lbrace0\rbrace),$$
this cone will have the collapsed point on the bottom (so it will be an "inverted" cone).

Question number one: I understood it right? This is the correct way to build a cone over a topological space $X$, in this case $S^1$?
Question number two: We collapsed one of the "ends" of the cylinder. The other points are deformed continuously? I mean, I have to understand this figure as if the points were connected between them and every point of the figure is suffering this deformation?

Now we have a cone. Let's consider the inverted one, $$\ (S^1\times [0,1])\  /\  (S^1\times \lbrace0\rbrace).$$

Question three: To build an homeomorphism from the cone to $D^2=\{x_1,x_2 \in \mathbb{R}^2 \mid x_1^2+x_2^2\leq1\}$, is the function $f(t,\theta)=t\cdot e^{i\theta}$ a good choice?

Thanks for your time!
 A: Question 1: Yes. You are right.
Q2: Hmm... You don't have to think of it that way. You are relying too much on visualization, which in my experience can lead to difficulties later on. To understand these constructions and concepts, it is best to have alternative ways of thinking about them. 
For instance, you can say that a cone is just the cylinder, but to have a continuous function from cone to somewhere, values must agree along the top ring; i.e. $f(x,1)=f(y,1)$ for all $x, y$ in your space. Continuity of $f$ will force nearby values to also be close to one another.
It can be that in some situations you require that up to half the height your cylinder be untouched, and only after that allow the points to "suffer the deformation." Talking about a topological space only, you don't have to impose any smoothness criterion, and up to homotopy equivalence, or even homeomorphism, a cone can be embedded in many weird shapes (upper hemisphere can be viewed as the cone over the circle.)
Q3. Please specify what direction your $f$ is taking? To check its continuity check the criteria I mentioned above. Intuitively, you would push down on the tip by your palm to flatten out the cone into the disk, so send the tip, all of the collapsed ring, to the cented of the disk. Level zeor ring, the bottom, sent to the out most circle of the disk.
