Birational morphism I have a question on rational and birational maps:
Is the map $$\mathbb{P}^1\rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:1)$$ rational? Birational? If birational what is its inverse?
 Same questions for map $$\mathbb{P}^1 \rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:0).$$
 My guess is that both aren't birational and that both are rational, but would like to hear another opinion. 
Thank you
 A: The first map you define is not even a map (except if you consider only the set structure and if the base field is $\Bbb F_2$). In $\Bbb P^1$, the points $[ x : y ]$ and $[ \lambda x : \lambda y ]$ are the same for all non zero $\lambda$ in the base field. So their image must be the same. But obviously all the $[\lambda x : \lambda y : 1 ]$ are not equal.
The second one is a morphism, defined everywhere, but it's not birational since its image is not dense in $\Bbb P^2$.
A: The first formula you give doesn't even define a set-theoretic map. 
The second defines a  morphism and  a fortiori a rational map.
But it  is  not birational because birational maps can  exist only between varieties of the same dimension.  
NB
Consider the open subset $U\subset \mathbb P^2$ consisting of points with coordinates $[x:y:1]$.
It is an affine variety (isomorphic to $\mathbb A^2$) and thus the only morphism 
$\mathbb{P}^1 \rightarrow \mathbb P^2$ with image included in $U$ are the constant ones $\mathbb{P}^1 \rightarrow \mathbb P^2: [x:y]\mapsto [a:b:1]$
