Question: Let rule $\varphi \colon (S^1 \times I)/(S^1 \times \{0\}) \rightarrow D^2$ defined by $\varphi([(z,t)]) = tz$ for all $[(z,t)] \in (S^1 \times I)/(S^1 \times \{0\})$. Show that $\varphi$ is a homeomorphism.
Could you give me some hints to solve this problem. Thank all!
From the point of view of category theory, this is the cone functor, $C(X)$, with $X=S^1\in \bf{Top}$, and $\varphi=C(id)$.
Think of polar coordinates. Every point in the disk, other than the origin, has a unique representation as $(\theta, r), \,\theta \in [0, 2\pi)$. The origin, on the other hand, can be represented by $(\theta,0)$, for any $\theta$. That's the reason for moding out by $S^1×\{0\}$, which identifies it to a point.
For injectivity, let $t_1z_1=t_2z_2$, for $t_1,t_2\in I$ and $z_1,z_2\in S^1$. Then $t_1=t_2$ and $z_1=z_2$.
For surjectivity, given $x\in D^2$, $x=tz=\varphi (z,t)$ for some $t\in I$ and $z\in S^1$. This is basically just uniqueness of the polar form, $re^{i\theta}$.
For continuity, let $(z,t)\to (z_0,t_0)$. Then $\varphi (z,t)=zt\to t_0z_0=\varphi (z_0,t_0)$. After all, it's the product of continuous functions, the projections onto each coordinate.
Similarly the inverse is continuous.
(So, you have a cylinder ($S^1×I$), and when you take the quotient by $S^1×\{0\}$, you identify one end to a point. This gives you a cone. But a cone is clearly homeomorphic to a disk.)