so I just started studying projective varieties (over algebraically closed fields) and I simply want to understand why $$V_{P_n}(T_i) \simeq P_{n-1}$$ whereas $V_{P_N}(T_i):= \{[z_0,..., z_n]\in P_n| z_i=0\}$ $i=0,...,n$.
So clearly I can get a morphism $P_{n-1}\mapsto V_{P_n}(T_i)$, $[z_o,...,z_{n-1}]\mapsto [z_0, .., 0, ..,z_{n-1} ]$ whereas the $0$ is at the i-th position, using that for every open set $U\subset P_n$, every prevariety Z and every morphism $\phi:\pi^{-1}(U)\mapsto Z$ such that $\phi(tz)= \phi(z)$ $\forall t\in K^{*}$ and $z\in\pi^{-1}(U)$, there exists a morphism $\psi:U\mapsto Z$ such that $\psi\circ\pi =\phi$. (whereas $\pi:\mathbb{K}^{n+1}\setminus\{0\}\mapsto P_n$).
Now I would like to construct the inverse morphism, which is given straightforwardly but I can't really get why it's also a morphism.
Does the same statement as above also hold for closed $A\subset P_n$ (i.e. the construction with the morphism $\psi:A\mapsto Z$)? Because thus I would immediately get what I want. More generally I don't really know how to construct isomorphism of closed sub varieties of $P_n$ (like, that $V_{P_2}(T_0T_1-T_2)\simeq P_1$).
I feel like at my stage, I really want to check all the details, but I don't really get it.