(Explicit) bijections between the affine $\mathbb{A}^n$ and the projective $\mathbb{P}^n$ space.

Question

I am referring to the complex case so let us fix $K=\mathbb{C}$ as base field.
I will consider the usual Zariski topology on the spaces.

I was reflecting on the existence of homeomorphisms between this two spaces, and in particulare I know the following two facts:

$\mathbb{A}^1$ and $\mathbb{P}^1$ are homeomorphic.
For $n\geq 2$ $\mathbb{A}^n$ and $\mathbb{P}^n$ are not homeomorphic

In particular, the first fact descends from the spaces having the same cardinality and having the cofinite topology, while in the second case we can reason on innntersection of hypersurfaces.

Anyhow I realized that:

1. I cannot explain why $\mathbb{A}^1$ and $\mathbb{P}^1$ have the same cardinality and a fortitori I cannot exhibit an explicit bijection between them
2. For the second problem to be of interest we must have a bijection between $\mathbb{A}^n$ and $\mathbb{P}^n$ but I cannot come up with one. In particular, the usual immersion of $\mathbb{A}^n$ in $\mathbb{P}^n$ is obviosuly not a bijection!
3. Do theese result change as we change the base field?

Can someone please help?

Answer 1

1. $\Bbb P^1_\Bbb C=\Bbb A^1_\Bbb C \cup \Bbb A^0_\Bbb C$ as sets. Since $|S_1\cup S_2|=\max(|S_1|,|S_2|)$ if one of $S_1$ or $S_2$ is infinite, we have the result. For an explicit bijection, define a map of sets $\Bbb A^1\to \Bbb P^1$ by sending $0\mapsto [0:1]$, $x\mapsto [1:x-1]$ for $x$ a positive integer, and $x\mapsto [1:x]$ for all other $x$.

2. We use the same trick as in 1. Write $\Bbb P^n_\Bbb C=\Bbb A^n_\Bbb C\cup \cdots \cdot \Bbb A^0_\Bbb C$. Thus both sides are of cardinality $|\Bbb C|$ and there exists a bijection between them. For an explicit bijection, the same "move it over" trick works: assume we've constructed a bijection between $\Bbb A^{n-1}$ and $\Bbb P^{n-1}$ by induction. Then we define a bijection between $\Bbb A^n$ and $\Bbb P^n$ by applying the bijection between points of the form $(x_1,\cdots,x_{n-1},0)$ and $[0:z_1:\cdots:z_n]$, and sending $(x_1,\cdots,x_n)$ to $[1:x_1:\cdots:x_n-1]$ for $x_n$ a positive integer and $[1:x_1:\cdots:x_n]$ otherwise.

3. Yes, for finite fields the point counts are explicitly different. For infinite fields, the same general principle applies, but one may have to use more interesting bijections than those constructed above (for instance, if the base field is $\overline{\Bbb F_2}$, then the field is infinite, but we don't have the integers around to move things over quite so nicely).