Question on Veronese map On page 10 of this note the author proved that the image of $\mathbb{P}^{n}$ under the Veronese map $\nu_{d}$ : $\mathbb{P}^{n}\longrightarrow \mathbb{P}^{m}$ is isomprphic to $\mathbb{P}^{n}$. 
In the proof, the author construct an inverse map : 

At each point of $\nu(\mathbb{P}^n)$, at least one of the coordinates indexed by the monomials $x_{0}^d,...,x_{n}^d$ must be nonzero. Let $U_i$ be the subset of $\nu(\mathbb{P}^n)$ where the coordinate indexed by $x_{i}^d$ is nonzero. The sets $U_0,...,U_n$ cover $\nu(\mathbb{P}^n)$  and we can define a map
$\varphi_{i}$ : $U_{i}\longrightarrow \mathbb{P}^n$ 
$z\mapsto (z_{(1,0,...,d-1,..0)}, z_{(0,1,...,d-1,..0)},..., z_{(0,0,...,d-1,..1)}$
That is, we send z to the (n+1)-tuple of its coordinates indexed by $x_0x_{i}^{d-1},...,x_nx_{i}^{d-1}$

My question is : $z\in U_i\subset \mathbb{P}^m$ then $z$ has $m$ coordinate. So, what did happen with the coordinate of $z$ ?
Could you please help me write down the map exactly like : $z=(z_1,...,z_m)\mapsto ...$
Thank for reading my question.
 A: It's a bit hard to write down explicitly in general, but the point is that you are picking out only $n+1$ of the original $m+1$ homogeneous coordinates of $z \in U_i$.
Let me illustrate with the example of the $3$-uple embedding of $\mathbb{P}^1$. Here, the Veronese map is $\nu_3 : \mathbb{P}^1 \rightarrow \mathbb{P}^3$ which sends $(X_0, X_1)$ to $(X_0^3, X_0^2 X_1, X_0 X_1^2, X_1^3)$.  Let $(Z_{(3,0)}, Z_{(2,1)}, Z_{(1,2)}, Z_{(0,3)})$ be the homogeneous coordinates on $\mathbb{P}^3$.  Then $U_0$ is the subset of $\mathbb{P}^3$ that consists of the points in $\nu_3(\mathbb{P}^1)$ where $Z_{(3,0)} \neq 0$.  For the inverse map on $U_0$, we send $(Z_{(3,0)}, Z_{(2,1)}, Z_{(1,2)}, Z_{(0,3)})$ to $(Z_{(3,0)}, Z_{(2,1)})$.  Similarly, $U_1$ is the subset of $\mathbb{P}^3$ that consists of the points in $\nu_3(\mathbb{P}^1)$ where $Z_{(0,3)} \neq 0$.  For the inverse map on $U_0$, we send $(Z_{(3,0)}, Z_{(2,1)}, Z_{(1,2)}, Z_{(0,3)})$ to $(Z_{(1,2)}, Z_{(0,3)})$.
Notice that these two inverse maps do not appear to be the same on their common overlap.  You have to use the fact that the points in $U_0 \cap U_1$ are in $\nu_3(\mathbb{P}^1)$ to prove that the two inverse maps do in fact agree on their common overlap, so they glue together to give a morphism $\nu_3(\mathbb{P}^1) \rightarrow \mathbb{P}^1$.  The general case is similar; one just has to be more careful with bookkeeping.
