Show that $\Bbb C P^n - \Bbb C P^{n-1} \cong \Bbb C^n.$
There is an obvious embedding of $\Bbb C P^{n-1}$ into $\Bbb C P^n$ given by $$[z_0,z_1, \cdots, z_{n-1}] \longmapsto [z_0,z_1, \cdots, z_{n-1}, 0].$$ In view of this embedding we can think of the elements of $\Bbb C P^n - \Bbb C P^{n-1}$ as the elements of $\Bbb C P^n$ of the form $[z_0,z_1, \cdots, z_n],$ where $z_n \neq 0.$ Since $\Bbb C P^n \cong (\Bbb C^{n+1} - 0) / (\Bbb C - 0)$ it follows that if $z_n \neq 0$ then $$[z_0,z_1, \cdots, z_n] = [z_0', z_1', \cdots, z_{n-1}', 1]$$ where $z_i' = \frac {z_i} {z_n},$ for $i = 0,1,2, \cdots, n-1.$ Consider the map $f : \Bbb C^n \longrightarrow \Bbb CP^n - \Bbb C P^{n-1}$ defined by $$(w_0, w_1, \cdots, w_{n-1}) \longmapsto [w_0, w_1, \cdots, w_{n-1}, 1].$$ Then $f = p \circ \iota,$ where $p : \Bbb C^{n+1} - 0 \longrightarrow (\Bbb C^{n+1} - 0)/ (\Bbb C - 0)$ is the quotient map whose image is $\Bbb C P^n - \Bbb C P^{n-1}$ and $\iota : \Bbb C^n \longrightarrow \Bbb C^{n+1} - 0$ is the inclusion given by $$(w_0, w_1, \cdots, w_{n-1}) \longmapsto (w_0, w_1, \cdots, w_{n-1},1).$$ So $f$ is continuous. Now consider the map $g : \Bbb C P^n - \Bbb C P^{n-1} \longrightarrow \Bbb C^n$ defined by $$[z_0,z_1, \cdots, z_n] \longmapsto (z_0', z_1', \cdots, z_{n-1}')$$ where $z_i' = \frac {z_i} {z_n},$ for $i = 0, 1, 2, \cdots, n-1.$ Then it is easy to see that $f$ and $g$ are inverses of each other. So if we can somehow show that $g$ is continuous then we are through. But I am unable to show that. Would anybody please help me in this regard?
Thanks in advance.
EDIT $:$ I am able to show that $$\Bbb C P^n - \Bbb C P^{n-1} \cong (\Bbb C^{n+1} - \Bbb C^n)/(\Bbb C - 0),$$ where the elements of $\Bbb C^n - \Bbb C^{n-1}$ are of the form $(z_0,z_1, \cdots, z_n),$ $z_i \in \Bbb C,$ for $i = 0, 1,2, \cdots, n$ and $z_n \neq 0.$ Now by the virtue of universal property of quotient topology, in order to prove the continuity of $g$ it is enough to show that the map $h : \Bbb C^{n+1} - \Bbb C^n \longrightarrow \Bbb C^n$ defined by $$(z_0,z_1, \cdots, z_n) \longmapsto (z_0', z_1', \cdots, z_{n-1}')$$ where $z_i' = \frac {z_i} {z_n},$ for $i = 0,1,2, \cdots, n-1,$ is a surjective continuous map with the fibres being the orbits of $\Bbb C^{n+1} - \Bbb C^n$ under the action of $\Bbb C - 0$ on $\Bbb C^{n+1} - \Bbb C^n$ by left (or right) multiplication which is somewhat an easy verification.
I am able to show that $$\Bbb C P^n - \Bbb C P^{n-1} \cong (\Bbb C^{n+1} - \Bbb C^n)/(\Bbb C - 0),$$ where the elements of $\Bbb C^{n + 1} - \Bbb C^n$ are of the form $(z_0,z_1, \cdots, z_n),$ $z_i \in \Bbb C,$ for $i = 0, 1,2, \cdots, n$ and $z_n \neq 0.$ Now by the virtue of universal property of quotient topology, in order to prove the continuity of $g$ it is enough to show that the map $h : \Bbb C^{n+1} - \Bbb C^n \longrightarrow \Bbb C^n$ defined by $$(z_0,z_1, \cdots, z_n) \longmapsto (z_0', z_1', \cdots, z_{n-1}')$$ where $z_i' = \frac {z_i} {z_n},$ for $i = 0,1,2, \cdots, n-1,$ is a surjective continuous map with the fibres being the orbits of $\Bbb C^{n+1} - \Bbb C^n$ under the action of $\Bbb C - 0$ on $\Bbb C^{n+1} - \Bbb C^n$ by left (or right) multiplication which is somewhat an easy verification.