A Veronese map is a morphism? My question is really simple, I'm beginning to study Algebraic Geometry and I'm still struggling to get the basic concepts. 
I would like to know if a Veronese map $v_{n,d}:\mathbb P^n\to \mathbb P^N$ is a morphism.
Thanks
 A: A morhism $\varphi:X \to Y$ between two projective varieties is a continous map that pulls back regular functions to regular functions. Explicitly, this means that if $U \subseteq Y$ is an open subset and $f:U \to k$ is a regular function, the composite $f \circ \varphi:\varphi^{-1}(U) \to k$ is also regular.
To check if a map of sets $\varphi:X \to \mathbb P^N$ is a morphism, you can do it locally. That is, let $U_i$ we the standard open cover of $\mathbb P^N$. Then each $U_i$ is isomorphic to $\mathbb A^{N}$ and has coordinate functions $\{ x_0,\cdots,x_n\}$. Then it is enough to check that each $x_i \circ \varphi:\varphi^{-1}(U_i) \to k$ is regular. (this is basically Lemma 3.6 in Chapter I of Hartshorne)
Now, lets look at your example in the case $n=1,d=2$ (for notational simplicity). Then the Veronese map takes the form $$v(p_0:p_1)=(p_0^2:p_0p_1:p_1^2)$$ for a point $(p_0:p_1) \in \mathbb P ^1$. The inverse images of the sets $U_0,U_1,U_2$ are $\mathbb P^1 \backslash\{p_0=0\}$, $\mathbb P^1 \backslash \{ (0:1),(1:0) \}$ and $\mathbb P^1 \backslash \{ p_1 = 0 \}$, respectively. 
Restricted to $U_0$, the Veronese map takes the form $$ v(1:p_1) = (1:p_1:p_1^2),$$  which is clearly regular, because it is a map defined by polynomials from one affine space to another.
The other charts are handled similarly, and it should not be difficult to generalize this to general $n$ and $d$.
