Does there exist a bijective mapping $f: \Delta\to\mathbb{C} $ Does there exist a bijective mapping $f: \Delta\to\mathbb{C} $?
Yet I have not found such example. Is it false (why?),please help me.
$ \Delta $ is the unit disk in $\mathbb {C}$.
 A: I would do this radius by radius. So in case we were talking about the open disk, all we’d need to do is get a nice map from $[0,1\rangle$ onto $[0,\infty\rangle$, like $x/(1-x)$. In case we were talking about the closed disk, you need a (discontinuous) map from $[0,1]$ onto $[0,1\rangle$, and follow it by the map you chose before. For this discontinuous map, one might send $1/n$ to $1/(n+1)$ for all $n\ge1$.
A: 1) I'll assume $\Delta$ is the unit circle first, since that's what you said in a comment above. Remove $1$ from the unit circle, and $(0,0)$ from the complex plane. Then take the polar coordinates. You see your question amounts to finding a bijection between $(0,2\pi)$ and $(0,+\infty)\times [0,2\pi)$. I will take for granted that for every interval which is not a singleton, there exists a bijection with $(0,1)$ (see here if you want to construct such bijections). Now the problem is equivalent to finding a bijection between $(0,1)$ and $(0,1)^2$.
Recall that every real number $x$ has a unique proper decimal expansion $x=x_0,x_1x_2\ldots$ By proper, I mean that the sequence of decimals is not eventually equal to $9$. Now the function 
$$
(0.x_1x_2\ldots,0.y_1y_2\ldots)\longmapsto 0.x_1y_1x_2y_2\ldots
$$
defines an injection of $(0,1)^2$ into $(0,1)$. It is easy to get an injection of $(0,1)$ into $(0,1)^2$. By Cantor-Schroeder-Bernstein, there exists a bijection between $(0,1)$ and $(0,1)^2$.
2) If $\Delta$ is the open unit disk now (which would make more sense notationwise), you can find a homeomorphism between $\Delta$ and $\mathbb{C}$. Note that you can't find a biholomorphism, due to Liouville. 
