Prove that, $\mathbb{C}P^{n} \setminus \mathbb{C}P^{n-1}$ is homeomorphic to $\mathbb{C}^{n}$ Prove that, $\mathbb{C}P^{n} 
\setminus \mathbb{C}P^{n-1}$ is homeomorphic to $\mathbb{C}^{n}$
To prove it, we first define, a map
$f:\mathbb{C}P^{n} 
\setminus \mathbb{C}P^{n-1}\to \mathbb{C}^{n}$ by
$[z_0,z_1,...,z_n]\to (w_0,..,w_{n-1})$ where, $w_i=\frac{z_i}{z_n}(z_n\ne 0).$
Clearly this map is bijective, we just need to prove that its a continuous map and inverse is also continuous.But, I am stuck here.Any idea?
 A: Your map is correct; you might want to write as a sentence, "we identify $\mathbb{P}^{n-1}$ as the subset of $\mathbb{P}^n$ where $z_n \neq 0$". As you say, your map is clearly invertible (you can lift ($w_0, \ldots, w_{n-1}$) to $[w_0: \ldots: w_{n-1}: 1]$ and every point in the subset of $\mathbb{P}^n$ where $z_n \neq 0$ has a unique representative of this form.)
For topology: you are viewing $\mathbb{P}^n$ as the quotient of $\mathbb{C}^{n+1} \setminus \{(0, \ldots, 0)\}$by the equivalence relation identifying $(a_0, \ldots, a_n) \sim (\lambda a_0, \ldots, \lambda a_n)$ for any $\lambda \neq 0$.
To show your map is continuous, you just have to extend it to a map
out of $\mathbb{C}^{n+1} \setminus q^{-1}([*:*:\ldots:*:0])$ (where $q$ denotes the quotient map) which respects equivalence classes; then it will be continuous by the universal property. (We are using a little lemma here: if $S \subset T$ is a saturated subset of a topological space with equivalence relation, then giving $S/\sim$ the quotient topology is the same as giving it the subspace topology from $T/\sim$).
To see that the inverse is continuous, note that the inverse  (which maps $(z_0, \ldots, z_{n-1})$ to $[z_0: \ldots, z_{n-1}:1]$) actually factors through the map to $\mathbb{C}^{n+1} \setminus \{(0, \ldots, 0)\}$ which just appends $1$ to every coordinate, and this map is clearly continuous.
